Lebesgue Integral

This part, we want to use conclusions from previous parts to extend the concept of integral.

The basic idea is, first define integral of simple function, then use convergence of simple function sequence to define integral of measurable function. Note that we first define them on non-negative functions. Then by making use of positive and negative part of function to extend to the whole functions.

Non-negative simple function integral

Lebesgue Integral of non-negative Simple Function

Assume \(f\) is a non-negative simple function defined on region measurable set \(D\), that is, there exists a partition \(\{E_i\}_{1\leq i\leq S}\), and non-negative real number set \(\{a_i\}_{1\leq i\leq S}\), such that

\[ f(x)=\sum_{i=1}^S a_i \chi_{E_i} (x),\quad x\in D. \]

Then its Lebesgue integral is defined by

\[ \int_{D}f(x)dx=\sum_{i=1}^S a_i m(E_i). \]

Properties

Invariance of Lebesgue Integral

Assume \(f\) and \(g\) are non-negative simple functions on measurable set \(D\), and satisfy \(f(x)=g(x), a.e.x \in D\), then their Lebesgue integrals on \(D\) are of the same.

Proof

Making use of the Partition of simple functions.

That is, let \(\{E_i\}_{1\leq i\leq N}\) and \(\{F_j\}_{1\leq j\leq M}\) to be the partition of \(f\) and \(g\) respectively, let non-negative number sets \(\{a_i\}\) and \(\{b_j\}\) to be the range of \(f\) and \(g\) respectively. Thus

\[ f(x)=\sum_{i=1}^N a_i \chi_{E_i}(x), \quad g(x)=\sum_{j=1}^M b_j \chi_{F_j}(x). \]

Once \(E_i\cap F_j\neq \varnothing\), we have \(a_i=b_j\).

Properties for L integral of non-negative simple function

Assume \(f\) and \(g\) are both simple functions on measurable set \(D\), then

(i) Monotonicity. If \(f(x)\leq g(x),a.e.x\in D\), then \(\int_D fdx\leq \int_D gdx\).

(ii) \(\int_D fdx \leq \max f(x)\cdot m(D)\). If \(m(D)=0\), then \(\int_D fdx=0\)

(iii) Linearity. If \(\lambda\) and \(\mu\) are two non-negative real number, then

\[ \int_D (\lambda f + \mu g)dx=\lambda \int_Dfdx+\mu \int_D gdx. \]

(iv) Interval additivity. If \(A, B\subset D\) and \(A\cap B=\varnothing\), then

\[ \int_{A\cup B}fdx=\int_A fdx+\int_B fdx. \]

Proof

(i) similar to proof in Invariance of Lebesgue Integral, just have \(a_i\leq b_j\).

(ii) easy to see.

(iii) Choose partition \(\{E_i\cap F_j\}_{1\leq i\leq N, 1\leq j\leq M}\).

(iv) For partition \(\{E_i\}_{1\leq i\leq N}\), see that

\[ E_i\cap (A\cup B)=(E_i\cap A)\cup (E_i\cap B) \]

and \((E_i\cap A)\cap (E_i\cap B)=\varnothing\).

The following lemma is a brigde between non-negative simple function and measurable function.

Lemma: Bridge

Assume \(g\) and \(\{f_n\}_{n\geq 1}\) are non-negative simple functions on measurable set \(E\), and satisfy

(i) \(\{f_n\}_{n\geq 1}\) are monotonically increasing \(a.e.x\in E\).

(ii) \(0\leq g(x)\leq \lim_{n\rightarrow \infty}f_n(x) (a.e.x\in E)\).

Then

\[ \int_{E}g(x)dx\leq \lim_{n\rightarrow \infty} \int_E f_n(x)dx. \]

Proof

Notice that \(\lim_{n\rightarrow \infty}f_n(x)\) would not be a simple function, so we could not use the monotonicity of Lebesgue integral for simple functions.

(i) \(g(x)\equiv c>0\).

\(\forall 0<\lambda<1\), define

\[ E_k:=\{f_n\geq \lambda c\}, \forall n\in \mathbb{N}^+. \]

It is easy to validify that \(E_1\subset E_2\subset \cdots\) and \(E=\bigcup_{n=1}^\infty E_n\). Thus

\[ \int_E f_1dx=\int_{E_n} f_ndx+\int_{E-E_n}f_ndx\geq \int_{E_n}f_ndx\geq \lambda c m(E_n) \]

let \(n\rightarrow \infty\) on both sides, we have

\[ \lim_{n\rightarrow \infty}\int_{E}f_ndx\geq \lambda c m(E)=\lambda\int_E gdx \]

at last we let \(\lambda\rightarrow 1\) and get the result.

(ii) Making use of the interval additivity.

Theorem for limit of monotonically increasing sequences

Assume \(\{f_n\}\) and \(\{g_n\}\) are non-negative simple functions on measurable set \(D\). If \(\forall x\in D\), \(\{f_n(x)\}\) and \(\{g_n(x)\}\) monotonically increase and converge to the same limit, then

\[ \lim_{n\rightarrow \infty}\int_D f_ndx = \lim_{n\rightarrow \infty}\int_D g_n dx. \]

Proof

The proof is to make use of the lemma above twice.

\(\forall n \geq 1\), \(\forall a.e. x\in D\), we have

\[ 0\leq g_n(x)\leq \lim_{k\rightarrow \infty}g_k(x)\leq \lim_{k\rightarrow \infty} f_k(x). \]

By Lemma, we have

\[ \int_D g_ndx\leq \lim_{k\rightarrow \infty}\int_D f_kdx. \]

Let \(n\rightarrow \infty\), we have \(\lim\limits_{n\rightarrow \infty}\int\limits_D g_ndx\leq \lim\limits_{k\rightarrow \infty}\int\limits_D f_kdx\). The opposite direction is of the same logic.

Non-negative measurable function integral

Lebesgue integral of measurable function

Assume \(f\) is a non-negative measurable function on measurable set \(D\). According to Approximation with Simple Function, there exists a non-negative function sequence \(\{f_n\}\) on \(D\), such that \(\forall x\in D\), \(\{f_n(x)\}\) monotonically increase and converges to \(f(x)\). Thus we define Lebesgue integral of \(f\) to be

\[ \int_D f dx =\lim_{n\rightarrow \infty }\int_D f_n dx. \]

When \(\int_D fdx<\infty\), we call \(f\) is L integrable on \(D\).

Example. Use definition of Lebesgue integral to calculate

\[ \int_{[0,\infty)} e^{-x}dx \]

Equivalent definition for Lebesgue integral of non-negative measurable function

Assume \(f(x)\) is a non-negative measurable function on \(E\), then

\[ \int_E fdx=\sup_{\varphi(x)\leq f(x)}\left\{\int_E \varphi(x) dx: \varphi(x) \text{ is a non-negative simple function}\right\}. \]

Or to be more specific, if we define \(\varPhi(f,E)\) is a set composed by non-negative simple function \(\varphi\) that satisfy \(\varphi\leq f\) on \(E\), then

\[ \lim_{k\rightarrow \infty}\int_E \varphi_k(x)dx=\sup_{\varphi\in \varPhi(f,E)}\int_E \varphi(x)dx. \]

Proof

• \(\leq\). Apparently, because \(\forall k\in \mathbb{N}^+\), we have \(\varphi_k\in \varPhi(f,E)\), so "\(\leq\)" holds naturally.

• \(\geq\). We know that \(\forall \varphi\in \varPhi(f,E)\), it satisfies \(\varphi(x)\leq f(x)=\lim_{k\rightarrow \infty}\varphi_k(x),\forall x\in E\), so by monotonicity

\[ \int_E \varphi(x)dx\leq\int_E\lim_{k\rightarrow \infty}\varphi_k(x)dx \]

by Theorem for limit of monotonically increasing sequences, we have

\[ \int_E \varphi(x)dx\leq\lim_{k\rightarrow \infty}\int_E \varphi_k dx \]

apply superior operation on both sides and we have "\(\geq\)" hold.

Properties

Properties for L integral of non-negative measurable function

Assume \(f\) is a non-negative measurable function on \(E\), then

(i) \(0\leq \int_E f(x)dx\leq \infty\).

(ii) Linearity. If \(\lambda\) and \(\mu\) are two non-negative real number, then

\[ \int_E (\lambda f + \mu g)dx=\lambda \int_E fdx+\mu \int_E gdx. \]

(iii) if \(A\subset E\) is also a measurable set, then

\[ \int_A f(x)dx=\int_E f(x)\chi_A(x)dx \]

(iv) if \(E=A\cup B, A\cap B=\varnothing\), then

\[ \int_E fdx=\int_Afdx+\int_B fdx. \]

(v) Monotonicity. If \(f(x)\leq g(x), \forall x\in E\), then

\[ \int_E f(x)dx\leq \int_E g(x)dx. \]

Proof

(iv) Using interval additivity of non-negative simple functions.

(v) similar to monotonicity for L integral of simple functions.

Example. Assume \(m(E)>0\), \(f\in L(E)\), \(f\geq 0\), prove: if

\[ \int_E fdx=0, \]

then

\[ f(x)=0,\quad a.e.x\in E. \]

Proof

Transfer problem into

\[ m(\{f(x)>0\})=0. \]

\(\forall n\geq 1\),

\[ \begin{align*} \int_E f(x)dx&\geq \int_{\{f\geq 1/n\}}f(x)dx\\ &\geq \int_{\{f\geq 1/n\}}\frac{1}{n}dx\\ &=\frac{1}{n}m(\{f\geq 1/n\}) \end{align*} \]

which means \(m(\{f\geq 1/n\})=0\), so

\[ \{f>0\}=\bigcup_{n=1}^\infty\{f\geq \frac{1}{n}\} =0, \]

which means \(f(x)=0, a.e.x\in E\).

Chebyshev Inequation

Assume \(f\) is a non-negative function on \(E\), then \(\forall \alpha>0\),

\[ m(\{f\geq \alpha\})\leq \frac{1}{\alpha}\int_E f(x)dx. \]

Proof

Let \(A=\{f\geq \alpha\}\), then

\[ \int_E fdx\geq \int_A fdx \geq \int_A \alpha dx = \alpha m(A). \]

Example. \(f_k(x)\geq 0\), \(f_k\in L(E)\). If

\[ \lim_{k\rightarrow \infty}\int_E f_k(x)dx=0, \]

then

\[ f_k\overset{m}{\rightarrow} 0. \]

Proof

\(\forall \sigma>0\), by Chebyshev Inequation

\[ \sigma \cdot m(\{|f_k-0|\geq \sigma\})\leq \int_{\{|f_k-0|\geq \sigma\}} f_kdx \leq \int_E f_kdx \rightarrow 0(k\rightarrow \infty) \]

which means \(m(\{|f_k-0|\geq \sigma\})\rightarrow 0\).

Example. Assume \(f\) is a non-negative function on \(\mathbb{R}\) with positive period \(T\), \(f\in L([0,T])\), prove:

\[ \lim_{x\rightarrow \infty}\frac{1}{x}\int_0^x f(t)dt=\frac{1}{T}\int_0^T f(t)dt. \]

Proof

For all \(k\geq 1\), we have

\[ \int_{kT}^{(k+1)T}f(t)dt=\int_0^T f(t+kT)dt=\int_0^Tf(t)dt. \]

So when \(kT\leq x\leq(k+1)T\), or \(\frac{1}{T}\geq \frac{k}{x}\geq \frac{k}{(k+1)T}\). Let \(k\rightarrow \infty\), we have the middle item

\[ \frac{k}{x}\rightarrow \frac{1}{T}. \]

so

\[ \begin{align} \nonumber\int_0^x f(t)dt&=\int_0^{kT}f(t)dt+\int_{kT}^x f(t)dt\\ \nonumber\int_0^xf(t) dt&=k\int_0^T f(t)dt+\int_{kT}^x f(t)dt\\ \Rightarrow \frac{1}{x}\int_0^xf(t)dt&=\frac{k}{x}\int_0^Tf(t)dt+\frac{1}{x}\int_{kT}^xf(t)dt\label{T-int} \end{align} \]

Since

\[ \left|\int_{kT}^x f(t)dt\right|\leq \int_{kT}^x |f(t)|dt\leq \int_{kT}^{(k+1)T} |f(t)|dt=\int_0^T|f(t)|dt=C \]

is bounded. So \(\frac{1}{x}\int_{kT}^xf(t)dt\rightarrow 0 (x\rightarrow \infty)\). So we let \(x\rightarrow \infty\) in equation \(\ref{T-int}\), we have the result.

Genaral measurable function integral

Lebesgue integral of Genaral measurable function

Assume \(f\) is a measurable fuunction on \(D\). \(\forall x\in D\), define positive and negative part of \(f\)

\[ f^+(x):=\max\{0,f(x)\},\quad f^-(x):=\max\{0,-f(x)\} \]

then it is easy to see that \(f^+\) and \(f^-\) are non-negative measurable functions and

\[ f(x)=f^+(x)-f^-(x),\quad |f(x)|=f^+(x)+f^-(x). \]

If either of \(f^+\) and \(f^-\) are \(\infty\) at the same time, then define Lebesgue integral of \(f\) to be

\[ \int_D f(x)dx=\int_D f^+(x)dx-\int_D f^-(x)dx. \]

when the above item is finite, we call \(f\) is Lebesgue integrable on \(D\), denoted as \(f\in L(D)\).

Equivalent condition for Lebesgue integrable

\(f\in L(D)\) iff \(|f|\in L(D)\).

When \(f\) is Lebesgue integrable, it apparently satisfies

\[ \left|\int_D fdx\right|\leq \int_D |f|dx \]

Proof

\(f\in L(D)\) iff \(f^+, f^-\in L(D)\) iff \(|f|\in L(D)\).

Properties

Properties for L integral of general measurable function

(i) If \(f\in L(E)\), then \(|f(x)|<+\infty, a.e.x\in E\).

(ii) If \(f\) is a generalized real function on zero-measure set \(A\), then \(\int_A fdx=0\).

(iii) If \(f\) and \(g\) are measurable functions on \(E\), and \(f(x)=g(x),a.e.x\in E\), then \(f\) and \(g\) are both L integrable or not integrable. When integrable, their values of integral are of the same.

(iv) Homotonicity & Linearity. If \(f, g\in L(E)\), \(a,b\in \mathbb{R}\), then \(af+bg\in L(E)\) and

\[ \int_E (af(x)+bg(x))dx=a\int_E f(x)dx+b\int_E g(x)dx. \]

(v) Interval additivity. If \(f\in L(E_1)\), \(f\in L(E_2)\), and \(E_1\cap E_2=\varnothing\), then \(f\in L(E_1\cup E_2)\) and

\[ \int_{E_1\cup E_2}fdx=\int_{E_1} fdx+\int_{E_2} fdx \]

Proof

(i) from equivalent definition.

(ii) easy to see.

(iii) \(f=g,a.e.x\in E\) \(\Rightarrow\) \(f^+=g^+, f^-=g^- a.e.x\in E\).

(iv) pay attention that \((af)^+\) has 3 different partition versions, i.e. \(a=0\), \(a>0\) and \(a<0\).

(v) Notice that \(|f+g|\leq |f|+|g|<\infty\), so \(f+g\in L(E)\).

Absolute Continuity of L integral

Assume \(f\in L(D)\). Then \(\forall \varepsilon>0\), \(\exists \delta>0\), such that \(\forall\) measurable set \(A\subset D\), so long as \(m(A)<\delta\), we have

\[ \left|\int_A f(x)dx\right|<\varepsilon. \]

Proof

Assume \(f\) is non-negative (discuss \(f^+\) and \(f^-\)). \(\forall \varepsilon>0\), \(\exists\) non-negative simple function \(\varphi\leq |f|\), such that

\[ \int_D |f|dx<\int_D \varphi dx+\frac{\varepsilon}{2}. \]

Because \(\varphi\) is bounded on \(D\), so \(\varphi(x)\leq M\). Choose \(\delta=\frac{\varepsilon}{2M}\), such that for all \(A\subset D\), \(m(A)<\delta\), we have

\[ \begin{align*} \left|\int_A fdx\right|&\leq \int_A|f|dx\\ &=\int_A (|f|-\varphi)dx + \int_A \varphi dx\\ &<\frac{\varepsilon}{2}+m(A)\cdot M<\varepsilon. \end{align*} \]

Levi Theorem

Levi Theorem

Assime \(f\) and \(\{f_n\}_{n\geq 1}\) are non-negative measurable function on \(D\), satisfy

\[ f_1(x)\leq f_2(x)\leq\cdots,\quad x\in D \]

and \(\lim\limits_{n\rightarrow \infty}f_n(x)=f(x)\), then

\[ \int_D fdx=\lim_{n\rightarrow \infty}\int_D f_n dx. \]

Proof version ⅠProof version Ⅱ

• \(\int_D fdx\geq\lim\limits_{n\rightarrow \infty}\int_D f_n dx\).

Notice that limit function \(f\) is also measurable, so its L integral has definition. Then since \(f_1\leq f_2\leq\cdots\leq f\), by monotonicity of L integral, we have \(\int_E f_n(x)dx\) monotonically increase, so its limit number \(\lim_{n\rightarrow \infty}\int_E f_n(x)dx\) has definition. By Monotonicity of L integral,

\[ \lim_{n\rightarrow \infty}\int_E f_ndx\leq \int_E fdx. \]

• \(\int_D fdx\leq\lim\limits_{n\rightarrow \infty}\int_D f_n dx\).

Choose a non-negative simple function \(\varphi(x)\), such that \(\varphi(x)\leq f(x),x\in E\). Let \(\lambda\in (0,1)\), denoted

\[ D_n=\{x\in D: f_n(x)\geq \lambda\varphi(x)\} \]

which is monotonically increasing, and \(\lim_{n\rightarrow \infty}D_n=D\). So

\[ \int_D f_ndx\geq \int_{D_n}f_ndx\geq \int_{D_n}\lambda \varphi(x)dx=\lambda\int_{D_n}\varphi(x)dx. \]

If we let \(n\rightarrow \infty\), the right side of inequation is a limit of integral region, i.e.

\[ \lim_{n\rightarrow \infty}\int_D f_ndx\geq \lambda\int_D \varphi(x)dx. \]

at last, let \(\lambda\rightarrow 1\), we have

\[ \lim_{n\rightarrow \infty}\int_D f_ndx\geq \int_D \varphi(x)dx. \]

by another definition of L integral of non-negative measurable function, we have

\[ \lim_{n\rightarrow \infty}\int_D f_ndx\geq \int_D f(x)dx. \]

Use another equvalent non-negative simple function sequence \(\{\psi_n\}\) of non-negative measurable function \(f_n\) to approximate \(f\).

Firstly, we have to make another simple function sequence useing its definition. That is, for every \(n\geq 1\), \(f_n\) is defined by sequence of non-negative simple function

\[ \{\varphi_k^{(n)}\}_{k\geq 1}, \]

which monotonically increase and converge to \(f_n\). Here, the method is a little tricky. For every \(k\geq 1\), define

\[ \psi_k(x):=\max\{\varphi_k^{(1)}(x),\varphi_k^{(2)}(x),\cdots,\varphi_k^{(k)}(x),\}, \quad x\in D. \]

So \(\psi_k(x)\) is also a non-negative simple function and

\[ \begin{align} \nonumber 0\leq \psi_1(x)\leq \psi_2(x)\leq\cdots\leq \psi_k(x)<\cdots,\quad x\in D.\\ \varphi_k^{(n)}(x)\leq \psi_k(x)\leq f_k(x),\quad 1\leq n\leq k,x\in D.\label{ineq1} \end{align} \]

Notice that inequation \(\ref{ineq1}\) use \(n\leq k\).

So by monotonicity of non-negative simple function, we have

\[ \begin{align} \int_D\varphi_k^{(n)}(x)\leq \int_D\psi_k(x)\leq \int_D f_k(x), \quad 1\leq n\leq k.\label{ineq2} \end{align} \]

In inequation \(\ref{ineq1}\), let \(k\rightarrow \infty\) firstly and let \(n\rightarrow \infty\), and we have

\[ \begin{align} \nonumber f_n(x)\leq \lim_{k\rightarrow \infty}\psi_k(x)&\leq f(x). \quad n\geq 1,x\in D.\\ \Rightarrow \lim_{k\rightarrow \infty}\psi_k(x)&=f(x),\quad x\in D.\label{eq1} \end{align} \]

Similarly, in inequation \(\ref{ineq2}\), let \(k\rightarrow \infty\) firstly and let \(n\rightarrow \infty\), and we have

\[ \begin{align} \nonumber \int_D f_n(x)dx=\lim_{k\rightarrow \infty}\int_D\varphi_k^{(n)}(x)&\leq \lim_{k\rightarrow \infty}\int_D\psi_k(x)\leq \lim_{k\rightarrow \infty}\int_D f_k(x), \quad n\geq 1.\\ \Rightarrow \lim_{k\rightarrow \infty}\int_D\psi_k(x)&=\lim_{n\rightarrow \infty}\int_D f_n(x)dx.\label{eq2} \end{align} \]

Equation \(\ref{eq1}\) means \(f\) could be approximated by \(\{\psi_K(x)\}\), which is just a nen-negative simple function sequence. Equation \(\ref{eq2}\) means

\[ \int_D f(x)dx\overset{\text{definition}}{=}\lim_{k\rightarrow \infty}\int_D\psi_k(x)=\lim_{n\rightarrow \infty}\int_D f_n(x)dx. \]

Example. Some examples which do not satisfy condition of Levi Theorem.

(i) \(f_k(x)=\chi_{[k,k+1]}(x),\quad E=\mathbb{R}\).

\(f_k(x)\rightarrow f(x)=0\), but

\[ \int_E f_kdx =1>0=\int_E fdx \]

(ii) \(f_k(x)=\frac{1}{k}\chi_{[0,k]}(x),\quad E=\mathbb{R}\).

\(f_k(x)\rightrightarrows f(x)=0\), but

\[ \int_E f_kdx =1>0=\int_E fdx \]

Example. Assume \(f\) is L integrable on \(E\), prove:

\[ \lim_{k\rightarrow \infty}k\cdot m(\{|f|>k\})=0. \]

Answer

Use \(\{|f|>k\}\) to narrow down the integral region.

Let

\[ f_k(x)=\begin{cases} f(x),\quad &|f|\leq k,\\ 0,\quad &|f|>k. \end{cases} \]

Then \(|f_k|\) \(\nearrow\) \(|f|\). By Levi Theorem,

\[ \begin{align*} \int_E |f|dx&=\lim_{k\rightarrow \infty} \int_E |f_k| dx\\ \Rightarrow \lim_{k\rightarrow \infty}\int_E(|f|-|f_k|)dx&=0 \end{align*} \]

While

\[ \begin{align*} \int_E(|f|-|f_k|)dx&= \int_{\{|f|\leq k\}}+\int_{\{|f|>k\}}(|f|-|f_k|)dx\\ &=\int_{\{f>k\}}|f|dx \\ &> k\cdot m(\{|f|>k\})\rightarrow 0 (k\rightarrow \infty). \end{align*} \]

There are two natural corollaries derived from Levi Theorem.

Corollary 1: Series form of Levi Theorem

Assume \(\{f_n\}_{n\geq 1}\) is a non-negative measurable function sequence, then

\[ \int_E \sum_{n=1}^\infty f_n(x)dx=\sum_{n=1}^\infty \int_E f_n(x)dx. \]

Proof

Let \(u_k(x)=\sum_{i=1}^k f_i(x)\), which satisfies the condition of Levi Theorem.

Corollary 2: Denumerable Additivity

Assume \(E\in \Omega\), \(E=\bigcup_{n=1}^\infty E_n\), where \(E_i\cap E_j=\varnothing\). If \(f\in L(E)\), \(f\geq 0\), then \(\forall k\in \mathbb{N}^+\), \(f\in L(E_k)\), and

\[ \int_E f(x)dx=\sum_{k=1}^\infty \int_{E_k} f(x)dx. \]

Proof

Let \(f_k(x)=f(x)\sum_{i=1}^k \chi_{E_i}(x)\) and use property (iii).

Fatou Lemma

Fatou Lemma

Assume \(\{f_n\}_{n\geq 1}\) is a non-negative measurable function sequence, then

\[ \int_D \underline{\lim}\limits_{n\rightarrow \infty} f_ndx\leq \underline{\lim}\limits_{n\rightarrow \infty} \int_D f_n dx. \]

Proof

From the definition of limit inferior of function, for every \(n\geq 1\), we could define

\[ g_n(x)=\inf_{k\geq n}f_k(x),\quad x\in D. \]

so \(\{g_n\}\) monotonically increase and converges to \(\underline{\lim}\limits_{n\rightarrow \infty}f_n(x)\). By Levi Theorem,

\[ \int_D \underline{\lim}\limits_{n\rightarrow \infty}f_n(x)dx=\lim_{n\rightarrow \infty}\int_D g_n(x)dx. \]

Since \(g_n(x)\leq f_n(x)\), so

\[ \begin{align*} \int_D \underline{\lim}\limits_{n\rightarrow \infty}f_n(x)dx&=\underline{\lim}\limits_{n\rightarrow \infty}\int_D g_n(x)dx\\ &\leq \underline{\lim}\limits_{n\rightarrow \infty}\int_D f_n(x)dx \end{align*} \]

This is a quite weak theorem, thus could be used in many fields.

Example. \(f_k(x)\geq 0\), \(f_k\in L(E)\), and \(f_k\rightarrow f, a.e.x\in E\). If

\[ \exists M>0, \int_E f_k(x)dx\leq M, \]

then

\[ f\in L(E). \]

Proof

\[ \begin{align*} \int_E fdx&=\int_{E_1}\lim_{k\rightarrow \infty}f_kdx\\ &\leq \lim_{k\rightarrow \infty}\int_{E_1}f_kdx \quad\text{by Fatou Lemma}\\ &\leq \lim_{k\rightarrow \infty}\int_{E}f_kdx<M \end{align*} \]

Note that limit and limit inferior are of the same here.

Example. Assume non-negative measurable function sequence \(\{f_k\}\) satisfies \(f_k\overset{m}{\rightarrow}f\), then prove hunmy

\[ \int_E fdx\leq \underline{\lim}\limits_{k\rightarrow \infty} f_kdx. \]

Proof

Get a subsequence \(\{f_{k_s}\}\) such that

\[ \int_E f_kdx=\lim_{s\rightarrow \infty}\int_{E} f_{k_s}dx \]

Since \(f_k\overset{m}{\rightarrow} f\), there exists a subsequence \(\{f_{k_{s_i}}(x)\}\rightarrow f(x), a.e.x\in E\), then by Fatou Lemma(1^st "\(\leq\)"), we have

\[ \begin{align*} \int_E fdx&=\int_E \lim_{i\rightarrow \infty} f_{k_{s_i}}dx\\ &\leq \lim_{i\rightarrow \infty} \int_E f_{k_{s_i}}dx\\ &\leq \lim_{s\rightarrow \infty} \int_E f_{k_{s}}dx\\ &=\underline{\lim}\limits_{k\rightarrow \infty} f_kdx. \end{align*} \]

Lebesgue's Dominated Convergence Theorem

Lebesgue's Dominated Convergence Theorem

Assume \(\{f_n\}_{n\geq 1}\) is a measurable function sequence \(a.e.\) on \(E\), \(f_n\rightarrow f, a.e.x\in E\). If there exists \(g\in L(E)\), such that \(\forall n\geq 1\), \(|f_n(x)|\leq g(x), a.e.x\in E\), then \(f\) and \(f_n\) is L integrable on \(E\) and

\[ \lim_{n\rightarrow \infty}\int_E f_ndx=\int_E f dx \]

where \(g\) is also called the controlling function.

Proof

Since \(|f_n|\leq g\), so its limit value \(|f|\leq g\), so \(g\in L(E)\) could induce \(f\) and \(f_n\) are L integrable.

Since \(|f_k-f|\leq |f_k|+|f|\leq 2g\), we could define

\[ g_k(x)=2g-|f_k-f|\geq 0, \]

then by Fatou Lemma

\[ \begin{align*} \int_E \underline{\lim}\limits_{k\rightarrow \infty}g_k(x)dx&\leq \underline{\lim}\limits_{k\rightarrow \infty}\int_E g_k(x)dx\\ \int_E \underline{\lim}\limits_{k\rightarrow \infty} (2g-|f_k-f|)dx&\leq \underline{\lim}\limits_{k\rightarrow \infty}\int_E (2g-|f_k-f|)dx\\ \int_E 2gdx &\leq \underline{\lim}\limits_{k\rightarrow \infty}\left[\int_E 2gdx - \int_E |f_k-f|dx\right]\\ \int_E 2gdx &\leq \int_E 2g dx + \underline{\lim}\limits_{k\rightarrow \infty} \left(-\int_E |f_k-f|dx\right)\\ 0&\leq -\overline{\lim}\limits_{k\rightarrow \infty}\int_E |f_k-f|dx\\ \overline{\lim}\limits_{k\rightarrow \infty}\int_E |f_k-f|dx&\leq 0 \end{align*} \]

So by relationship of limit superior and inferior

\[ 0\leq \underline{\lim}\limits_{k\rightarrow \infty}|f_k-f|dx\leq \overline{\lim}\limits_{k\rightarrow \infty}|f_k-f|dx\leq 0 \]

which gives

\[ \lim_{k\rightarrow \infty}|f_k-f|dx=0 \]

Since

\[ \left|\int_E f_kdx-\int_E fdx\right|=\left|\int_E (f_k-f)dx\right|\leq \int_E |f_k-f|dx\rightarrow 0(k\rightarrow \infty) \]

and we are done.

Another form of Dominated Convergence Theorem

Assume \(\{f_n\}_{n\geq 1}\) is a measurable function sequence \(a.e.\) on \(E\), \(f_n\overset{m}{\rightarrow} f, a.e.x\in E\). If there exists \(g\in L(E)\), such that \(\forall n\geq 1\), \(|f_n(x)|\leq g(x), a.e.x\in E\), then \(f\) and \(f_n\) is L integrable on \(E\) and

\[ \lim_{n\rightarrow \infty}\int_E f_ndx=\int_E f dx \]

Proof

Use contradiction and the above Dominated Convergence Theorem.

Let \(a_k=\int_E f_kdx\), \(a=\int_E fdx\). Assume \(a_k\nrightarrow a\), then \(\exists \{k_j\}\), such that the convergence of its subsequence

\[ \lim_{j\rightarrow \infty}\int_E f_{k_j}dx=A\neq a. \]

we want to find a point that contradicts. Since \(f_k\overset{m}{\rightarrow}f\), we have \(f_{k_j}\overset{m}{\rightarrow}f\). By Riesz Theorem, we have

\[ f_{k_{j_i}}\rightarrow f, a.e.x\in E \]

Combined with \(|f_{k_{j_i}}|\leq g\), by Dominated Convergence Theorem, we have

\[ \lim_{i\rightarrow \infty}\int_E f_{k_{j_i}}dx=\int_E fdx=A \]

while \(a_{k_j}\rightarrow A\neq a\), we have \(a_{k_{j_i}}\rightarrow A\neq a\), which contradicts!

Applications of LDCT

The following come some applications of LDCT.

Corollary 1: Term-by-term Integration Theorem

Assume \(\{f_n\}\) are measurable functions on \(D\), satisfies

\[ \sum\limits_{n=1}^\infty \int\limits_E |f_n|dx<\infty, \]

then \(f:=\sum\limits_{n=1}^\infty f_n\) converges absolutely \(a.e.\) on \(D\), and

\[ \int_E fdx = \sum_{n=1}^\infty \int_E f_ndx \]

Proof

Consider define \(F_k(x):=\sum\limits_{j=1}^\infty f_j,k\in \mathbb{N}^+\), then \(g=\sum\limits_{k=1}^\infty |f_k|\).

Apply LDCT to \(F_k\).

Corollary 2: theorem for limit of integral region

Assume \(f\) is a measurable function on \(E\), \(E_1\subset E_2\subset \cdots\) is a sequence of subsets of \(E\), which satisfy \(E=\bigcup_{k=1}^\infty E_k\). If \(f\in L(E)\), then

\[ \int_E fdx=\lim_{k\rightarrow \infty}\int_{E_k}fdx. \]

Proof

Let \(f_k(x):=f(x)\chi_{E_k}(x)\), so \(\lim_{k\rightarrow \infty}f_k=f(x)\) and \(|f_k|\leq |f|\in L(E)\), by LDCT, we have

\[ \begin{align*} \int_{E}fdx&=\lim_{k\rightarrow \infty}\int_{E}f_kdx\\ &=\lim_{k\rightarrow \infty}\int_{E}f(x)\chi_{E_k}dx\\ &=\lim_{k\rightarrow \infty}\int_{E_k}f(x)dx \end{align*} \]

Example. Calculate

\[ \lim_{n\rightarrow \infty}\int_{(0,+\infty)}\frac{n\sqrt{x}}{1+n^2x^2}\sin^{24}(nx)dx \]

Answer

Let \(f_n(x)=\frac{n\sqrt{x}}{1+n^2x^2}\sin^{24}(nx)\), \(f(x)=0\). Here we have to discuss a controlling function.

• \(x\in(0,1]\). Notice

\[ \left|\frac{n\sqrt{x}}{1+n^2x^2}\sin^{24}(nx)\right|\leq \left|\frac{n\sqrt{x}}{1+n^2x^2}\right|\leq \frac{n\sqrt{x}}{2nx}=\frac{1}{\sqrt{x}}. \]

whose L integral is finite on \((0,1]\), so Choose \(g(x)=\frac{1}{2\sqrt{x}}\).

• \(x\in (1,+\infty)\). Notice

\[ \left|\frac{n\sqrt{x}}{1+n^2x^2}\right|\leq \frac{n\sqrt{x}}{n^2x^2}\leq \frac{1}{x^{3/2}}. \]

whose L integral is finite on \((1,+\infty)\), so choose \(g(x)=\frac{1}{x^{3/2}}\).

Thus consider

\[ g(x)=\begin{cases} \frac{1}{2\sqrt{x}},\quad x\in(0,1]\\ \frac{1}{x^{3/2}},\quad x\in (1,+\infty). \end{cases} \]

So \(f_n(x)\) satisfy condition of LDCT, meaning

\[ \lim_{n\rightarrow \infty}\int_{(0,+\infty)}\frac{n\sqrt{x}}{1+n^2x^2}\sin^{24}(nx)dx=\int_{(0,+\infty)}0dx=0. \]

Corollary 3: Theorem of denumberable additivity

Assume \(a\) is a measurable function on \(E\), \(\{E_k\}_{k\geq 1}\) is a sequence of subsets of \(E\), which satisfies \(E=\bigcup_{k} E_k\) and \(E_i\cap E_j=\varnothing, \forall i\neq j\). If \(f\in L(E)\), then

\[ \int_E fdx=\sum_{k=1}^\infty \int_{E_k}fdx. \]

Proof

Construct another sequence of sets \(\{\tilde{E_k}\}\) as

\[ \tilde{E_k}:=\bigcup_{j=1}^kE_j, \]

which monotonically increase and \(\lim\limits_{k\rightarrow \infty}\tilde{E_k}=E\), so by Corollary 2: theorem for limit of integral region, we have

\[ \begin{align*} \int_E fdx&=\lim_{k\rightarrow \infty}\int_{\tilde{E_k}} fdx\\ &=\lim_{k\rightarrow \infty}\int_{\bigcup_{j=1}^kE_j} fdx\\ &=\lim_{k\rightarrow \infty}\sum_{j=1}^k \int_{E_j}fdx\\ &=\sum_{j=1}^\infty \int_{E_j}fdx. \end{align*} \]

Notice derivative is also a limit operation, so here comes the following theorem.

Corollary 4: Parameter derivatives of L integral

Assume \(f(x,y)\) is a function on \(E\times I\), where \(E\in \Omega(\mathbb{R}^n)\), \(I\subset \mathbb{R}\) is an interval. If

(i) \(\forall y\in I\), \(f(x,y)\in L(E)\) with respect to \(x\),

(ii) \(\forall x\in E\), \(f(x,y)\in C(I)\) with respect to \(y\),

(iii) \(\exists 0\leq F(x)\in L(E)\), such that

\[ \left|\frac{\partial }{\partial y}f(x,y)\right|\leq F(x),\quad \forall (x,y)\in E\times I, \]

then

\[ \frac{\partial}{\partial y}\int_E f(x,y)dx=\int_E \frac{\partial}{\partial y}f(x,y)dx,\quad \forall y\in I. \]

Proof

\(\forall y\in I\), \(\exists h_k>0\), s.t. \(y+h_k\in I\), then define

\[ g_k(x,y):=\frac{f(x,y+h_k)-f(x,y)}{h_k}. \]

which is apprarently a measurable function, and its limit function is \(g(x)=\frac{\partial }{\partial y}f(x,y)\). We have to find an appropriate controlling function \(G(x)\) to control \(g_k\). Notice that by mean value theorem

\[ g_k(x,y)=\left|\frac{\partial }{\partial y}f(x,y+\theta_k h_k)\right|\leq F(x),\quad \forall x\in E, \text{ where }\theta_k\in (0,1). \]

So by LDCT, we have

\[ \begin{align*} \lim_{k\rightarrow \infty}\frac{1}{h_k}\left[\int_E f(x,y+h_k)dx-\int_E f(x,y)dx\right]&=\lim_{k\rightarrow \infty}\int_{E} g_k(x,y)dx=\int_{E}\lim_{k\rightarrow \infty} g_k(x,y)dx\\ \frac{\partial}{\partial y}\int_E f(x,y)dx&=\int_E \frac{\partial}{\partial y}f(x,y)dx \end{align*} \]

Assume \(f\in L[a,b]\), prove

\[ \frac{d}{dx}\int_{[a,b]}f(y)\sin(xy)dy=\int_{[a,b]}yf(y)\cos(xy)dy. \]

Answer

Easy to see that \(F(x)=f(y)\sin(xy)\in L([a,b])\) with respect to \(y\) and \(F(x)\in C(I)\) with respect to \(x\). Notice that

\[ \left|\frac{\partial}{\partial x}F(x)\right|=|yf(y)\cos(xy)|\leq |yf(y)| \in L([a,b]\times I). \]

So by Theorem for Parameter derivatives of L integral, we are done.

Corollary 5: Approximation with functions

Assume \(f\in L([a,b])\), prove: \(\forall \varepsilon>0\), \(\exists\)

(i) bounded measurable function \(g\), such that \(\int_a^b|f-g|dx<\varepsilon\).

(ii) continuous function \(h\), such that \(\int_a^b|f-h|dx<\varepsilon\).

(iii) polynomial \(P\), such that \(\int_a^b|f-P|dx<\varepsilon\).

(iv) step function \(S\), such that \(\int_a^b|f-S|dx<\varepsilon\).

Proof

(i) Let \(f_k(x)=f\cdot \chi_{\{|f|\leq k\}}(x)\). Since

\[ |f_k(x)-f(x)|\leq |f(x)|\in L([a,b]) \]

and \(\lim_{k\rightarrow\infty}|f_k-f|=0,a.e.x\in [a,b]\). By LDCT we have

\[ \lim_{k\rightarrow \infty}\int_a^b |f-f_k|dx=0. \]

So there exists \(K\), when \(k\geq K\), we have

\[ \int_a^b |f-f_k|dx<\varepsilon. \]

Let \(g=f_{k_0}\) for some \(k_0\geq K\).

(ii) By (i), there exists bounded measurable function \(g\) such that \(\int_a^b|f-g|dx<\frac{\varepsilon}{2}\). Let \(|g(x)|\leq M\). To get a continuous function, we could use Luzin theorem, there exists continuous function \(h\) such that \(m(\{h\neq g\})<\frac{\varepsilon}{4M}\) and \(\max|h(x)|\leq M\), then

\[ \begin{align*} \int_a^b |f-h|dx&\leq \int_a^b|f-g|dx+\int_a^b|g-h|dx\\ &\leq \frac{\varepsilon}{2}+2M\cdot m(\{g\neq h\})\\ &<\frac{\varepsilon}{2}+2M\cdot \frac{\varepsilon}{4M}=\varepsilon. \end{align*} \]

Riemann Integral & Lebesgue Integral

Deduction

Assume \(f\) is bounded on \([a,b]\), and we have a partition \(\{x_k\}_{0\leq k\leq n}\) on \([a,b]\)

\[ \Delta_k: a=x_0<x_1<\cdots<x_n=b. \]

Define \(M_j=\sup_{x\in [a,b]}\{f(x)\}\), \(n_j=\inf_{x\in [a,b]}\{f(x)\}\), and define two function

\[ \begin{align*} M_{\Delta_k} (x)=\sum_{j=1}^n M_j \chi_{[x_{j-1},x_j]}(x)\\ m_{\Delta_k} (x)=\sum_{j=1}^n m_j \chi_{[x_{j-1},x_j]}(x) \end{align*} \]

which apparently is measurable function (simple function), and we have

\[ \begin{align*} m_{\Delta_k}(x)\leq f(x)\leq M_{\Delta_k} (x) \end{align*} \]

So we have an important relationship

\[ \begin{align*} \int_{[a,b]}M_{\Delta_k}(x)dx=\sum_{j=1}^n M_j m([x_{j-1},x_j])=\sum_{j=1}^n M_j \Delta x_k = \overline{S}_{\Delta_k}(f)\\ \int_{[a,b]}m_{\Delta_k}(x)dx=\sum_{j=1}^n m_j m([x_{j-1},x_j])=\sum_{j=1}^n m_j \Delta x_k = \underline{S}_{\Delta_k}(f) \end{align*} \]

Since \(\Delta_k\) is not denumerable, so we choose to let \(\Delta_k\) to be an equi-distant partition of \([a,b]\) with distance \(N_k=2^k\):

\[ \Delta_k: a=x_0^{(k)}<x_1^{(k)}<\cdots<x_{N_k}^{(k)}=b \]

and \(\Delta_{k+1}\) is a refinement of \(\Delta_{k}\), \(\Vert\Delta_k\Vert\rightarrow 0 (k\rightarrow \infty)\).

Since \(\int_{[a,b]}m_{\Delta_k}(x)dx\) does not increase, \(\int_{[a,b]}M_{\Delta_k}(x)dx\) does not decrease, we have their limit

\[ \begin{align*} \lim_{k\rightarrow\infty}\int_{[a,b]}M_{\Delta_k}(x)dx=\lim_{k\rightarrow \infty}\overline{S}_{\Delta_k}(f) = \inf \overline{S}_{\Delta_k}(f)\\ \lim_{k\rightarrow\infty}\int_{[a,b]}m_{\Delta_k}(x)dx=\lim_{k\rightarrow \infty}\underline{S}_{\Delta_k}(f) = \inf \underline{S}_{\Delta_k}(f) \end{align*} \]

So from what we have known in Riemann integral,

\[ \begin{equation} f\in R[a,b]\Leftrightarrow \lim_{k\rightarrow\infty}\int_{[a,b]}m_{\Delta_k}(x)dx=\lim_{k\rightarrow\infty}\int_{[a,b]}M_{\Delta_k}(x)dx \label{R-L-1} \end{equation} \]

Deduction 2

Thus we have to use integral limit theorem. That is, define Baire upper and lower function \(M(x)=\lim\limits_{k\rightarrow \infty}M_{\Delta_k}(x)\), \(m(x)=\lim\limits_{k\rightarrow \infty}m_{\Delta_k}(x)\), which is bounded, satisfying condition of LDCT, i.e.

\[ \begin{align*} \lim_{k\rightarrow \infty}\int_{[a,b]} M_{\Delta_k}(x)dx=\int_{[a,b]}m(x)dx\\ \lim_{k\rightarrow \infty}\int_{[a,b]} m_{\Delta_k}(x)dx=\int_{[a,b]}m(x)dx. \end{align*} \]

So equivalence \(\ref{R-L-1}\) becomes

\[ \begin{align*} f\in R[a,b]&\Leftrightarrow \int_{[a,b]}m(x)dx=\int_{[a,b]}M(x)dx\\ &\Leftrightarrow \int_{[a,b]}[M(x)-m(x)]dx=0\\ &\Leftrightarrow M(x)=m(x), a.e.x\in [a,b]\quad \text{by } \end{align*} \]

The last Equivalence is due to Example.

Note that about Barie functions, we have a theorem

Baire Theorem

Assume \(f\) is bounded on \([a,b]\), \(x_0\in [a,b]\), then \(f\) is continuous at \(x_0\), iff \(M(x_0)=m(x_0)\).

Proof

• "\(\Rightarrow\)".

\(f\) is continuous at \(x_0\), then \(\forall \varepsilon>0\), \(\exists \delta>0\), such that

\[ |f(x)-f(x_0)|< \varepsilon,\quad \forall x\in O_\delta(x_0) \]

that is, \(f(x_0)-\varepsilon<f(x)<f(x_0)+\varepsilon\)。 There exists sufficient large \(k\) such that

\[ f(x_0)-\varepsilon<m_{\Delta_k}(x)<M_{\Delta_k}(x)<f(x_0)+\varepsilon. \]

• "\(\Leftarrow\)".

Finally we have the following theorem.

Riemann integral and Lebesgue integral

If \(f\in R[a,b]\), then \(f\in L[a,b]\), and

\[ (R)\int_a^b fdx=(L)\int_{[a,b]}fdx \]

Proof

Sufficient & Necessary condition for Riemann integrable

Assume \(f\) is bounded on \([a,b]\), then \(f\in R[a,b]\), iff \(f\in C(a.e.x\in [a,b])\).

Proof

• "\(\Rightarrow\)".

Example.

(i) Riemann function

\[ R(x)=\begin{cases}\frac{1}{q},\quad x=\frac{p}{q}\\ 0,\quad x\in \mathbb{Q}^c \end{cases}. \]

is Riemann integrable.

(ii) Characteristic function of Cantor set \(\chi_C(x)\) is Riemann integrable.

Generalized Riemann integral

Because genaralized Riemann integral is a limit in a sense of Cauchy, we could not say generalized Riemann integral could be extended into Lebesgue integral.

Example. \(f(x)=\sin x/x\), then their generalized integral

\[ \int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}. \]

but

\[ \int_0^\infty \left|\frac{\sin x}{x}\right|=\infty. \]

so \(f\notin L([0,\infty))\).

In another words, Lebesgue integral is an absolute convergent intergal. So we have the following relation regarding generalized integral.

Generalized Riemann integral of non-negative function

(i) Assume \(f(x)\geq 0\), if

\[ (R)\int_0^\infty fdx=\lim_{A\rightarrow \infty}\int_0^A fdx \]

converges, then \(f\in L((0,\infty))\) and

\[ (L)\int_{(0,\infty)}fdx=(R)\int_0^\infty fdx. \]

Proof

(i) First we prove that \(f\in L((0,\infty))\). Notice that

\[ f\in R([a,b])\Rightarrow f\in L([a,b]) \]

let \(b\rightarrow \infty\), we get \(f\in L((0,\infty))\).

Then we prove the equation. Notice that

\[ (R)\int_0^\infty fdx=\lim_{k\rightarrow \infty}(R)\int_0^kfdx=\lim_{k\rightarrow \infty}(L)\int_{(0,k)}fdx=\lim_{k\rightarrow \infty}\int_{(0,\infty)}f(x)\chi_{(0,k)}(x)dx \]

Define \(f_k(x)=f(x)\chi_{(0,k)}(x)\), then \(f_k \nearrow\), satisfy condition of Levi Theorem, so

\[ (L)\int_{(0,\infty)}fdx=\lim_{k\rightarrow \infty}(L)\int_{(0,\infty)}f_kdx=(R)\int_0^\infty fdx \]

Corollary: Generalized Riemann integral

Assume \(f\) is bounded on finite intervals and

\[ (R)\int_0^\infty |f|dx=\lim_{A\rightarrow \infty}\int_0^A |f|dx \]

converges, then \(f\in L((0,\infty))\) and

\[ (L)\int_{(0,\infty)}fdx=(R)\int_0^\infty fdx. \]

Note that we can prove in a similar way that

\[ (L)\int_{(-\infty,+\infty)}fdx=(R)\int_{-\infty}^{+\infty} fdx. \]

Proof

This function might not be non-negative, so we should use LDCT. That is, Define \(f_k(x)=f(x)\chi_{(0,k)}(x)\), which converges to \(f(x)\), and

\[ |f_k(x)|\leq |f(x)| \]

while the latter one is L integrable, so by LDCT, we have \(f\in L((0,\infty))\) and

\[ (L)\int_{(0,\infty)}f dx=\lim_{k\rightarrow \infty}(L)\int_{(0,k)}f_kdx=\lim_{k\rightarrow \infty} (R)\int_0^k f_kdx = \lim_{k\rightarrow \infty} (R)\int_0^k fdx = \int_0^\infty fdx \]

Theoretically speaking, the Riemann integral of conditional convergent function could not be extended to Lebesgue integral.

Multiple Lebesgue integral

This part we aim to find whether the following equation holds

\[ \begin{align} \int_{\mathbb{R}^p\times \mathbb{R}^q}f(x,y)dxdy=\int_{\mathbb{R}^q}\left[\int_{\mathbb{R}^p}f(x,y)dx\right]dy\label{multiple-integral} \end{align} \]

where \(x\in \mathbb{R}^p\), \(y\in \mathbb{R}^q\).

Firstly, we consider characteristic functions.

Deduction 1

Assume \(f=\chi_E\), then equation \(\ref{multiple-integral}\) becomes

\[ \begin{align} m(E)=\int_{\mathbb{R}^p\times \mathbb{R}^q}\chi_E(x,y) dxdy=\int_{\mathbb{R}^q}\left[\int_{\mathbb{R}^p}\chi_E (x,y)dx\right]dy.\label{mid-equation-integral} \end{align} \]

Definition of Cross Section

Assume \(E\in \Omega(\mathbb{R}^{p+q})\), \(\forall y\in \mathbb{R}^q\), then define

\[ E^y:=\{x\in \mathbb{R}^p| (x,y)\in E\} \]

the cross section of \(E\) at \(y\).

Deduction 2

With the above definition, check that \(\chi_E(x,y)=1\Leftrightarrow (x,y)\in E\), i.e. \(x\in \{x\in \mathbb{R}^p: (x,y)\in E\}=E^y\). So we could have an equivalent relation

\[ \chi_E(x,y)=\chi_{E^y}(x). \]

Thus equation \(\ref{mid-equation-integral}\) becomes

\[ m(E)=\int_{\mathbb{R}^q}\left[\int_{\mathbb{R}^p}\chi_{E^y} (x)dx\right]dy=\int_{\mathbb{R}^q}m(E^y)dy. \]

Before we begin proof, we have to give some properties of the cross section.

Properties about cross section

Assume \(A,B,\{E_k\}\) are subsets of \(\mathbb{R}^{p+q}\), \(\forall y\in \mathbb{R}^q\), then

(i) If \(A\subset B\), then \(A^y\subset B^y\).

(ii) \((\bigcup_k E_k)^y=\bigcup_k E_k^y\), \((\bigcap_k E_k)^y=\bigcap_k E_k^y\).

(iii) \((A-B)^y=A^y-B^y\).

(iv) If \(E_k \nearrow A\), then \((E_k)^y\nearrow A^y\); If \(E_k \searrow A\), then \((E_k)^y\searrow A^y\)

Theorem for Cross Section

Assume \(E\subset \mathbb{R}^{p+q}\) is measurable, then

(i) \(\forall a.e.y\in \mathbb{R}^q\), \(E^y\) is measurable with respect to \(y\).

(ii) \(m(E^y)\) is a non-negative function on \(\mathbb{R}^q\) and

\[ \int_{\mathbb{E}^q}m(E^y)dy=m(E). \]

Proof

• (i) \(E=I\times J\), where \(I\subset \mathbb{R}^p\), \(J\subset \mathbb{R}^q\) are half-open cubes.

So \(\forall a.e.y\in \mathbb{R}^q\), we have

\[ E^y=\begin{cases} I,\quad y\in J\\ \varnothing,\quad y\notin J \end{cases} \]

meaning \(E^y\) is measurable. Its measure (a function of \(y\))

\[ m(E^y)=m(I)\cdot \chi_{J}(y) \]

So by Measure of direct product (the 3^rd "="), its integral

\[ \begin{align*} \int_{\mathbb{R}^q}m(E^y)dy&=\int_{\mathbb{R}^q}m(I)\chi_J (y)dy\\ &=m(I)m(J)\\ &=m(I\times J)=m(E). \end{align*} \]

• (ii) \(E\) is a bounded open set on \(\mathbb{R}^{p+q}\). So by configuration of open sets, we have

\[ E=\bigcup_{k=1}^\infty I_k, \quad I_k\subset \mathbb{R}^{p+q}, \text{ & } E_i\cap E_j=\varnothing. \]

So

\[ E^y=\left(\bigcup_{k=1}^\infty I_k\right)^y = \bigcup_{k=1}^\infty I_k^y \]

which is measurable. Then by denumerable additivity

\[ \begin{align} m(E^y)=\sum_{k=0}^\infty m(I_k^y)\label{Ey} \end{align} \]

is a non-negative measurable function. Thus by Series form of Levi Theorem (the 3^rd "=") and equation \(\ref{Ey}\) (the 4^th "=")

\[ \begin{align*} m(E)&=\sum_{k=1}^\infty m(I_k)\\ &=\sum_{k=1}^\infty \int_{\mathbb{R}^q} m(I_k^y)dy\quad\text{by (i)}\\ &=\int_{\mathbb{R}^q} \sum_{k=1}^\infty m(I_k^y)dy \\ &=\int_{\mathbb{R}^q} m(E^y)dy \end{align*} \]

• (iii) \(E\) is a bounded \(G_\delta\) set.

That is, \(E=\bigcap_{k=1}^\infty G_k\), where \(G_k\) are mutually disjoint open sets. Here we let \(G_k\) monotonically decrease (increase makes no sense for intersection), i.e. \(G_1\supset G_2\supset \cdots\), so the cross section

\[ E^y=\bigcap_{k=1}^\infty G_k^y, \quad G_1^y\supset G_2^y\supset\cdots \]

so by limit operation of measure, its measure (a function of \(y\))

\[ m(E^y)=m(\lim_{k\rightarrow \infty}G_k^y)=\lim_{k\rightarrow \infty}m(G_k^y) \]

meaning it could be expressed as a form of limit of measurable functions. So we could use LDCT, if we check that

\[ |m(G_k^y)|\leq |m[(I\times J)^y]|=m(I)\chi_J(x)\in L(\mathbb{R}^q). \]

So by LDCT,

\[ \begin{align*} \int_{\mathbb{R}^q}m(E^y)dy&=\lim_{k\rightarrow \infty}\int_{\mathbb{R}^q}m(G_k^y)dy\\ &=\lim_{k\rightarrow \infty}m(G_k)\quad \text{by (ii)}\\ &=m(E) \end{align*} \]

(iv) \(E\) is a zeo-measure set in \(\mathbb{R}^{p+q}\).

So by [Approximation me]

(v) \(E\) is a bounded measurable set.

(vi) \(E\) is a general measurable set.

Tonelli Theorem

Then we consider \(f\) is a measurable function.

Tonelli Theorem

Assume \(f(x,y)\) is a non-negative measurable function with \((x,y)\in \mathbb{R}^p\times \mathbb{R}^q\), then

(i) \(\forall a.e.y\in \mathbb{R}^q\), \(f(x,y)\), as a function of \(x\in \mathbb{R}^q\) is non-negative measurable.

(ii) \(F(y):=\int_{\mathbb{R}^p}f(x,y)dx\) is non-negative measurable on \(\mathbb{R}^q\), and equation \(\ref{multiple-integral}\) holds.

Proof

(i) \(f\) is a simple function, which holds naturally, bacause Linearity of Lebesgue integral.

(ii) \(f\) is a non-negative measurable funcion. Notice that

\[ f(x)=\lim_{k\rightarrow \infty}\varphi_k(x,y) \]

where \(\varphi_k\) are non-negative simple functions, and monotonically increase.

Example. Assume \(f\) is a measurable function on \(E\), then

\[ \int_E |f|dx=\int_{(0,\infty)}d_f(\alpha)d\alpha \]

where \(d_f(\alpha)=m(\{f>\alpha\})\).

Proof

\[ \begin{align*} \int_{(0,\infty)}d_f(\alpha)d\alpha&=\int_{(0,\infty)}m(\{f>\alpha\})d\alpha\\ &=\int_{0,\infty}\left[\int_E \chi_{\{f>\alpha\}}(x)dx\right]d\alpha\\ &=\int_E dx\int_{(0,\infty)} \left[\int_E \chi_{\{f>\alpha\}}(x)d\alpha\right]\quad \text{by Tonelli Theorem}\\ &=\int_E dx \int_{(0,|f|)} \left[\int_E \chi_{\{f>\alpha\}}(x)d\alpha\right]\\ &=\int_E |f|dx. \end{align*} \]

Fubini Theorem

Fubini Theorem

Assume \(f(x,y)\) is a L integrable function with \((x,y)\in \mathbb{R}^p\times \mathbb{R}^q\), then

(i) \(\forall a.e.y\in \mathbb{R}^q\), \(f(x,y)\), as a function of \(x\in \mathbb{R}^q\) is L integrable.

(ii) \(F(y):=\int_{\mathbb{R}^p}f(x,y)dx\) is L integrable on \(\mathbb{R}^q\), and equation \(\ref{multiple-integral}\) holds.

Proof

Notice that \(f(x,y)=f(x,y)^+-f(x,y)^-\), which are non-negative measurable function. So by Tonelli Theorem, we are done.

Example. Calculate L integral \((0< a < b)\)

\[ I=\int_{(0,\infty)}\frac{\sin x}{x}[e^{-ax}-e^{-bx}]dx \]

Answer

Notice that

\[ I=\int_0^\infty \sin x dx\int_{a}^b e^{-xy}dy \]

Check whether \(f=\sin x e^{-xy}\) is L integrable, i.e. for \(|f|\),

\[ \begin{align*} \int_{(0,\infty)}\int_a^b |f|dxdy&\leq \int_{(0,\infty)}\int_a^b e^{-xy}dxdy\\ &=\int_a^b dy\int_{(0,\infty)}e^{-xy}dx\quad\text{by Tonelli Theorem}\\ &=\int_a^b \frac{1}{x}dx=\ln (b/a)<\infty \end{align*} \]

which is L integrable. So by Fubini Theorem,

\[ \begin{align*} I&=\int_a^b dy\int_{(0,\infty)}e^{-xy}\sin x dx\\ &=\int_a^b \frac{1}{1+y^2}dy=\arctan b-\arctan a \end{align*} \]

Example. prove

\[ \int_{(0,\infty)^2}e^{-xy}\sin x\sin ydxdy=\int_{(0,\infty)}\sin y dy\int_{(0,\infty)}e^{-xy}\sin xdx=\int_{(0,\infty)}\sin x dx\int_{(0,\infty)}e^{-xy}\sin ydy \]