Measurable Function

Functions discussed in this chapter are defined on measurable sets, whose value could be generalized real number, i.e. \(\mathbb{R}\) including \(\pm \infty\).

Definition of Measurable Function

Assume function \(f\) has defined on measurable set \(D\). If \(\forall \alpha\in \mathbb{R}\), set

\[ \{f>\alpha\}=D(f>\alpha):=\{x\in D: f(x)> \alpha\} \]

is also measurable, then we call \(f\) is a measurable function on \(D\).

Example. Continuous function on an interval \(D\subset \mathbb{R}\) is a measurable function.

Proof

• For an open interval \(D\), this is easy.

• For an arbitrary interval \(D\), the original image of \({f>\alpha}\) would differ from open sets with at most \(2\) points.

Characteristic Function

Assume \(A\subset X\), then define a characteristic function \(A\) as

\[ \chi_A(x)=\begin{cases}1,\quad x\in A\\ 0, \quad x\in X\backslash A \end{cases} \]

Measurability of Characteristic Function

For measurable set \(D\), denote \(\chi_D\) as its characteristic function, then

\[ \{\chi_D>\alpha\}=\begin{cases} \varnothing,\quad \alpha\geq 1\\ D,\quad 0\leq\alpha<1 \\ \mathbb{R},\quad \alpha<0 \end{cases} \]

So we could prove \(\chi_D\) is measurable function.

Properties

Properties of Measurable Function

Assume \(f(x)\) is measurable on \(E\), then the following sets are all measureable.

(i) \(\{f\leq \alpha\}\) (ii) \(\{f\geq \alpha\}\)

(iii) \(\{f<\alpha\}\) (iv) \(\{f=\alpha\}\)

(v) \(\{f<+\infty\}\) (vi) \(\{f=+\infty\}\)

(vii) \(\{f>-\infty\}\) (viii) \(\{f=-\infty\}\)

Proof

• (i) \(=E-\{f>\alpha\}\)

• (ii) \(=\bigcap_{k=1}^\infty\{f>\alpha-\frac{1}{k}\}\) (length decrease, so we choose inferior)

• (iii) using (ii) with same logic of proving (i)

• (iv) using (i) amd (ii)

• (v) \(=\bigcup_{k=1}^\infty\{f<k\}\) (length increase, so we choose superior)

• (vi) use (v) with same logic of proving (i)

• (vii) same logic of proving (i)

• (viii) use (vii) with same logic of proving (i).

The above (i), (ii) and (iii) are equivalent definition of measurable function, but (iv) is not.

Example. \(f\) is measurable on measurable set \(D\), iff \(\forall r\in \mathbb{Q}\), \(\{f>r\}\) is measurable.

Proof

• "\(\Rightarrow\)". Let \(\forall \alpha\), \(\alpha=r\).

• "\(\Leftarrow\)". Notice that \(\forall \alpha, \exists \{r_n\}\), such that \(r_n\in \mathbb{R}\), and \(r_1>r_2>\cdots\), and \(r_n\rightarrow \alpha\), so

\[ \{f>\alpha\}=\bigcup_{n=1}^\infty\{f>r_n\} \]

which means \(f\) is measurable.

When \(\{f=r\}\) is measurable, we could not deduce the same result. Bacause let \(A\) to be a non-measurable set, define

\[ f=\begin{cases} \sqrt{3},\quad x\in A\\ \sqrt{2},\quad x\notin A \end{cases} \]

then \(\forall r\in \mathbb{Q}\), \(\{f=r\}=\varnothing\) is measurable, but let \(\alpha=1.5\),

\[ \{f>1.5\}=A \]

is not measurable, so \(f\) is not measurable.

Region Extension

(i) Assume generalized real function \(f(x)\) is defined on \(D_1\cup D_2\). If \(f(x)\) is measurable on \(D_1\) and \(D_2\), then \(f(x)\) is also measurable on \(D_1\cup D_2\).

(ii) Assume \(f(x)\) is measurable on \(D\). If \(A\subset D\) is also measurable, then \(f(x)\) is also measurable on \(A\).

Proof

(i) \(\{x\in D_1\cup D_2:f>\alpha\}=\{x\in D_1: f>\alpha\}\cup \{x\in D_2: f>\alpha\}\) and by measure operation.

(ii) \(\{x\in A: f>\alpha\}=\{x\in D: f>\alpha\}\cap A\) and by measure operation.

Lemma of functions

Assume \(f(x)\), \(g(x)\) are measurable on \(E\), then \(\{f>g\}\) is also measurable.

Proof

\(\forall \varepsilon_r=\frac{1}{r},r\in \mathbb{N}^+\), we have

\[ \{f>g\} = \bigcup_{r=1}^\infty\{f> \varepsilon_r\}\cap\{\varepsilon_r >g\} \]

so \(\{f>g\}\) is measurable.

Basic Operations

Basic Operations of Measurable Function

Assume \(f(x)\), \(g(x)\) are measurable function on \(D\), then the following functions are measurable.

(i) \(cf(x) (c\in \mathbb{R})\). (ii) \(f(x)+g(x)\). (iii) \(f(x)\cdot g(x)\).

Proof

(i) move \(c\) to the side of \(\alpha\).

(ii) same logic with Lemma.

(iii) use \(4f(x)\cdot g(x)=[f(x)+g(x)]^2-[f(x)-g(x)]^2\). You only need to consider whether \(f^2(x)\) are measurable, which is easy to see. Notice that \(\forall\alpha\geq 0\),

\[ \{f^2>\alpha\}=\{f>\sqrt{\alpha}\}\cup \{f<-\sqrt{\alpha}\} \]

is measurable, and \(\forall \alpha<0\),

\[ \{f^2>\alpha\}=D \]

is measurable.

Example. Assume \(f\) is defined on measurable set \(D\), prove: if \(f^2\) is measurable on \(D\) and \(\{f>0\}\) is measurable, then \(f\) is measurable on \(D\).

Proof

• \(\forall \alpha\geq 0\),

\[ \{f>\alpha\}=\{f^2>\alpha^2\}\cap \{f>0\} \]

is measurable.

• \(\forall \alpha<0\),

\[ \{f>\alpha\}=\{f>0\}\cup \{\alpha<f\leq 0\}=\{f>0\}\cup \left[\{f^2<\alpha^2\}\cap \{f\leq 0\}\right] \]

is measurable.

Example. Assume \(f\) is a measurable function on \(D\), prove: for all open set \(G\subset \mathbb{R}\) and closed set \(F\subset \mathbb{R}\),

\[ f^{-1}(G),\quad f^{-1}(F) \]

are measurable.

Proof

Known from configuration of set, \(\forall\) open set \(G\subset \mathbb{R}\), there exists \(\{a_n\}\) and \(\{b_n\}\), such that \(a_n<b_n\), and \(G=\bigcup\limits_{n=1}^\infty(a_n,b_n)\), so

\[ f^{-1}(G)=\bigcup_{n=1}^\infty\{a_n<f<b_n\} \]

is measurable. Notice that \(f^{-1}(F)=f^{-1}(\mathbb{R}-F^c)=D-f^{-1}(F^c)\) is also measurable.

Limit superior & limit inferior

Limit superior & limit inferior

Sequence of Function.

Similarly, given a sequence of function \(\{f_n(x)\}_{n\geq 1}\) we have limit superior

\[ \overline{\lim}\limits_{n\rightarrow \infty}f_n=\inf_{n\geq 1} \{\sup_{k\geq n}\{f_k(x)\}\} \]

and limit inferior

\[ \underline{\lim}\limits_{n\rightarrow \infty}f_n=\sup_{n\geq 1} \{\inf_{k\geq n}\{f_k(x)\}\} \]

Theorem for measurability of Limit

Assume \(\{f_n(x)\}_{n\geq 1}\) is a sequence of measurable functions on measurable set \(D\), then the following functions are all measurable.

\[ \begin{align} \sup_{n\geq 1}f_n(x),\quad \inf_{n\geq 1}f_n(x)\label{supinf-function}\\ \overline{\lim}\limits_{n\rightarrow \infty}f_n(x),\quad \underline{\lim}\limits_{n\rightarrow \infty}f_n(x) \label{limit-supinf-function} \end{align} \]

Proof

• For expression \(\ref{supinf-function}\), we can write it as

\[ \left\{\sup_{n\geq 1}f_n(x)>\alpha\right\}=\bigcup_n^\infty\{f_n(x)>\alpha\} \]

where the latter one is measurable.

• For expression \(\ref{limit-supinf-function}\), we could use conclusion from \(\ref{supinf-function}\).

Example. Assume \(f_n\) is a measurable function on \(D\). Prove: Set

\[ \{x\in D: f_n \text{ converges}.\} \]

is measurable.

Proof

\(f_n\) converges, iff its limit superior and inferior are of the same. So we define

\[ F(x)=\overline{\lim}\limits_{n\rightarrow \infty}f_n,\quad f(x)=\underline{\lim}\limits_{n\rightarrow \infty}f_n, \]

both of which are measurable by Theorem for limit. Then consider

\[ \{F=f\}=D-\{F>f\} \]

since the latter is measurable according to Lemma of functions, \(\{F=f\}\) is measurable.

Almost Everywhere

Definition of Almost Everywhere of a Proposition

Assume \(D\) is measurable, and proposition \(P(x)\) is related with points in \(D\). If there exists a zero-measure set \(E\subset D\), such that \(P(x)\) holds on \(D\backslash E\), then we call \(P(x)\) holds almost everywhere on \(D\).

We denote this statement as

\[ P(x) \ a.e.x \in D. \]

Approximation

Approximation with Simple Function

Simple Function

Assume \(f\) is a function on measurable set \(D\). If \(f(D)\) is composed of finite points \(\{a_k\}_{k=1}^n\), and sets \(E_k=\{f=a_k\} (k=1,2,\cdots, n)\) are all measurable, then we call \(f\) is a simple function on \(D\).

Apparently, here \(f(x)\) could be expressed by finite linear combination of characteristic functions, i.e.

\[ f(x)=\sum_{k=1}^n a_k \chi_{E_k}(x) \]

Approximate measurable function with sequence of simple functions

If \(f(x)\) is measurable on \(D\), then there exists a sequence of simple functions \(\{f_k\}_{k=1}^\infty\), such that for every point \(x\in D\), \(\{f_k\}_{k\geq 1}\) converges to \(f(x)\). Specially, if

(i) \(f\) is non-negative, then \(\{f_k\}_{k\geq 1}\) is monotonically increasing.

(ii) \(f\) is bounded, then \(\{f_k\}_{k\geq 1}\) converges uniformly to \(f(x)\).

HintsProof

Use a special construction.

Formulate a sequence of function which converges to \(f(x)\). This is a little tricky.

Define

\[ f_n(x)=\begin{cases} n,\quad f(x)>n\\ \frac{k-1}{2^n},\quad \frac{k-1}{2^n}\leq f(x)<\frac{k}{2^n}, k=-n2^n+1,-n2^n\cdots,n2^n\\ -n,\quad f(x)<n \end{cases} \]

Look the following image for intuitional impression.

Now we show that \(f_n(x)\) corvergese to \(f(x)\). For fixed point \(x\in D\), if

(i) \(f(x)=\infty\). Then \(f_n(x)=n\rightarrow \infty=f(x)\). The same for \(f(x)=-\infty\).

(ii) \(-\infty<f(x)<+\infty\). Then for sufficient large number \(n\), there exists unique \(k_n\) such that \(-n 2^n+1\leq k_n \leq n2^n\) and

\[ \frac{k_n-1}{2^n}\leq f(x)\leq \frac{k_n}{2^n} \]

which means \(f_n(x)=\frac{k_n-1}{2^n}\), satisfying

\[ 0\leq f(x)-f_n(x)\leq \frac{1}{2^n}. \]

Let \(n\rightarrow \infty\), we have \(f_n(x)\) converges to \(f(x)\).

• Specially, if \(f(x)\) is non-negative, for \(f(x)=+\infty\), then \(f_n(x)=n\) is monotonically increasing.

For \(f(x)<\infty\), if \(f(x)<n\), we have \(f(x)=\frac{k-1}{2^n}\) for some \(k\) such that \(\frac{k-1}{2^n}<f(x)<\frac{k}{2^n}\), then for \(f_{n+1}(x)\), we partition this interval into two parts, which are

\[ \left[\frac{2k-2}{2^{n+1}},\frac{2k-1}{2^{n+1}}\right), \left[\frac{2k-1}{2^{n+1}}, \frac{2k}{2^{n+1}}\right) \]

the former part \(f_{n+1}(x)=f_n(x)=\frac{k-1}{2^n}\), while the latter part gives \(f_{n+1}(x)=\frac{2k-1}{2^{n+1}}>f_n(x)\), so it is monotonically increasing.

for \(\infty>f(x)\geq n\), this is similar to \(\infty\).

• Specailly, if \(f(x)\) is bounded, denote its upper bound as \(M\), so when \(n>M\), we have \(|f(x)-f_n(x)|<\frac{1}{2^n}\) on all \(x\in D\), which means \(f_n(x)\rightrightarrows f(x)\).

Example. Assume \(f\) is differentiable on \(\mathbb{R}\), prove \(f'(x)\) is measurable.

Proof

\(f\) is differentiable, so \(f\) and \(f(x+\frac{1}{n})\) is continuous, thus measurable. So

\[ f'(x)=\lim_{n\rightarrow \infty} \frac{f(x+\frac{1}{n})-f(x)}{\frac{1}{n}}\overset{\Delta}{=}f_n(x) \]

\(f_n\) is measurable, so \(f'\) is also measurable.

Uniform Convergnce a.e.

Lemma: use language of Set to describe convergence

Assume \(\{f_k(x)\}_{k\geq 1}\), \(f(x)\) are measurable function on \(a.e.x \in E\), and \(m(E)<\infty\). If \(f_k(x)\rightarrow f(x), a.e.x \in E\), then \(\forall \varepsilon>0\), let \(E_k(\varepsilon)=\{x\in E: |f_k(x)-f(x)|\geq \varepsilon\}\), then we have

\[ \lim_{j\rightarrow \infty} m\left(\bigcup_{k=j}^\infty E_k(\varepsilon)\right)=0. \]

Proof

We have the following equivalent statement for point-wise convergence.

\[ \begin{align} \nonumber &f_k(x)\rightarrow f(x), a.e.x \in E \\ \nonumber \Leftrightarrow & \forall \varepsilon>0, \exists N>0, \forall k\geq N, |f_k(x)-f(x)|<\varepsilon\\ \nonumber \Leftrightarrow & \forall \varepsilon>0, \exists N>0, x\in \bigcap_{k=N}^\infty \{x\in E: |f_k(x)-f(x)|<\varepsilon \}\\ \Leftrightarrow & \forall \varepsilon>0, x\in \bigcup_{N=1}^\infty \bigcap_{k=N}^\infty \{x\in E: |f_k(x)-f(x)|<\varepsilon \} \label{equiv-pointwise-convergence}\\ \end{align} \]

Note that we could have an equivalent statement at the last

\[ x\in \bigcap_{\varepsilon>0}\bigcup_{N=1}^\infty \bigcap_{k=N}^\infty \{x\in E: |f_k(x)-f(x)|<\varepsilon \} \]

but \(\bigcap_{\varepsilon>0}\) is not denumerable, so the in the following proof, we have to choose a denumerable set like \(\mathbb{Q}\) or \(\frac{1}{r}\).

Now we have a equivalent statement of \(f_k(x)\rightarrow f(x), a.e.x\in E\), that is, expression \(\ref{equiv-pointwise-convergence}\). Then its opponent proposition \(f_k(x)\nrightarrow f(x)\) could be expressed by

\[ \exists \varepsilon>0, x\in \bigcap_{N=1}^\infty \bigcup_{k=N}^\infty \{x\in E: |f_k(x)-f(x)|\geq \varepsilon \} \]

That is,

\[ \begin{align} \{x\in E: f_k(x)\nrightarrow f(x)\}=\bigcup_{\varepsilon>0}\bigcap_{N=1}^\infty \bigcup_{k=N}^\infty E_k(\varepsilon) \label{non-convergence} \end{align} \]

which has zero-measure according to \(a.e.x\in E\), so by limit operations in measure, we have

\[ \lim_{N\rightarrow \infty}m\left(\bigcup_{k=N}^\infty E_k(\varepsilon)\right)=m\left(\lim_{N\rightarrow \infty} \bigcup_{k=N}^\infty E_k(\varepsilon)\right)=m\left( \bigcap_{N=1}^\infty \bigcup_{k=N}^\infty E_k(\varepsilon)\right)=0. \]

The first equation holds because \(E_k(\varepsilon)\) is measurable (\(f_k, f\) are all measurable functions) and \(\bigcup_{k=N}^\infty E_k(\varepsilon)\) is monotonically decreasing when \(N\rightarrow \infty\). The last equation holds because the left side of equation \(\ref{non-convergence}\) has zero-measure (which means the item to be unioned should have zero-measure).

Egoroff Theorem

Assume \(\{f_k(x)\}_{k\geq 1}\) is a measurable function which is bounded on \(a.e.x\in E\), and \(m(E)<\infty\). If \(f_k(x)\rightarrow f(x), a.e.x\in E\), then \(\forall \delta>0\), \(\exists \text{ measurable set } E_\delta\subset E\), such that \(m(E_\delta)\leq \delta\) and

\[ f_k(x)\rightrightarrows f(x), \quad \forall x\in E-E_\delta. \]

Proof

Choose \(\varepsilon=\frac{1}{r}\) and define

\[ E_k\left(\frac{1}{r}\right)=\{x\in E: |f_k(x)-f(x)|\geq \frac{1}{r}\}, \]

then by condition of the theorem and Lemma, \(\forall \delta>0\), \(\forall r\in \mathbb{R}\), \(\exists N_r\), \(\forall N \geq N_r\),

\[ m\left[\bigcup_{k=N}^\infty E_k\left(\frac{1}{r}\right)\right]<\frac{\delta}{2^r}. \]

Then we have to sum up there measure. That is, let \(E_\delta=\bigcup_{r=1}^\infty \bigcup_{k=N}^\infty E_k\left(\frac{1}{r}\right)\), which satisfies

\[ m(E_\delta)\leq \sum_{r=1}^\infty m\left[\bigcup_{k=N}^\infty E_k\left(\frac{1}{r}\right)\right]<\delta. \]

Check that

\[ E\backslash E_\delta=\bigcap_{r=1}^\infty \bigcap_{k=N}^\infty E_k^c\left(\frac{1}{r}\right)=\bigcap_{r=1}^\infty \bigcap_{k=N}^\infty \{x\in E: |f_k(x)-f(x)|< \frac{1}{r}\} \]

Notice that \(f_k\rightrightarrows f, a.e.x\in A\), iff \(\sup_{x\in A}|f_k(x)-f(x)|\rightarrow 0 (k\rightarrow \infty)\). If we let \(A=E\backslash E_\delta\), this holds apparently.

Note that the condition \(m(E)<\infty\) in the above theorem could not be removed.

Example. \(f_n(x)=x^n\) converges to \(f(x)=\begin{cases}0,\quad 0<x<1 \\ 1,\quad x=1 \end{cases}\) point-wisely, but not uniformly. If we remove a sufficiently small set \((1-\delta,1]\), then \(f_n(x)\rightrightarrows f(x)\) uniformly.

Extension of Measurable Function

The following theorem is the main result of this part.

Lusin Theorem

Assume \(f(x)\) is a measurable function \(a.e.\) on \(E\), then \(\forall \delta>0\), \(\exists\) closed set \(F\subset E\), s.t. \(m(E\backslash F)<\delta\), and \(f(x)\) is a continuous function on \(F\).

HintsProof

Using Theorem in Mathematical Analysis for uniform convergent function, that is, the convergent function of uniformly convergent function sequence is continuous(Here notice we have to cut off a closed set from \(E\), which might be infinite). So we have to make use of Egoroff Theorem.

(i) \(f\) is a simple function on \(E\).

From conclusion in Property of Simple Function, there exists measurable sets \(E_i(i=1,2,\cdots,N)\), such that \(E=\bigcup_{i=1}^N E_n\), \(E_i\cap E_j=\varnothing\), and

\[ f(x)=\sum_{i=1}^N \alpha_i \chi_{E_i}(x) \]

From Theorem of Approximation for measurable sets, \(\forall \delta>0\), because \(E_i\) is measurable, \(\exists\) closed set \(F_i\subset E_i\), such that \(m(E_i\backslash F_i)<\delta/N\), and \(\chi_{E_i}(x)\in C(F_i)\), or to be more specific, is constant on \(F_i\). So from \(F:=\bigcup_{i=1}^N F_i\subset E=\bigcup_{i=1}^N E_i\), we have

\[ \begin{align*} m(E\backslash F)&=m\left[\left(\bigcup_{i=1}^N E_i \right)\backslash \left(\bigcup_{i=1}^N F_i\right)\right]\\ &\leq m\left[\bigcup_{i=1}^N \left( E_i\backslash F_i\right)\right]\\ &\leq \sum_{i=1}^N m(E_i\backslash F_i)<\delta \end{align*} \]

(ii) \(m(E)<\infty\).

From conclusion in Approximation with simple function, there exists a sequence of simple function \(\{\psi_k\}_{k\geq 1}\) such that \(\psi_k(x)\rightarrow f(x)\).

Then since \(m(E)<\infty\), we could use Egoroff Theorem for the above function sequence. That is, \(\forall \delta>0\), \(\exists \text{ measurable function }E_\delta\subset\), such that \(m(E\backslash E_\delta)<\delta/2\) and

\[ \psi_k(x)\rightrightarrows f(x),\quad \forall x\in E\backslash E_\delta. \]

The following method is a little tricky. \(\forall k\), from (i) we could know for simple function \(\psi_k(x)\), there exists closed set \(F_k\subset E\backslash E_\delta\), such that \(m((E\backslash E_\delta)\backslash F_k)<\delta/2^{k+1}\) and \(\psi_k(x)\in C(F_k)\). Define \(F=\bigcap_{k=1}^\infty F_k\) which is still a closed set.

Now we have to prove that the part we remove is small enough. That is, since

\[ E\backslash F=E\backslash \left(\bigcap_{k=1}^\infty E_k\right) \subset [E_\delta] \cup \left[(E\backslash E_\delta)\backslash \left(\bigcap_{k=1}^\infty E_k\right)\right] \]

so

\[ \begin{align*} m\left[E\backslash F\right]&\leq m\left[E_\delta\right] + m\left[(E\backslash E_\delta)\backslash \left(\bigcap_{k=1}^\infty E_k\right)\right]\\ &\leq \frac{\delta}{2}+m\left[\bigcup_{k=1}^\infty \left((E\backslash E_\delta)\backslash F_k\right)\right]\\ &\leq \frac{\delta}{2} + \sum_{k=1}^\infty m\left[\left((E\backslash E_\delta)\backslash F_k\right)\right]\\ &\leq \frac{\delta}{2} + \frac{\delta}{2}=\delta \end{align*} \]

Thus, \(\psi_k(x)\) uniformly converges to \(f(x)\) on \(F\), by Theorem in Mathematical Analysis for uniform convergent function, meaning \(f(x)\in C(F)\).

(iii) Last situation. \(m(E)=\infty\).

We have to transform this problem into (ii), that is, use finite-measure set to intersect.

Define \(E_k:=E\cap \{x\in E: k-1 \leq |x|< k\}\). Then \(E=\bigcup_{k=1}^\infty E_k\), and \(E_i\cap E_j=\varnothing\).

So \(\forall k\), there exists closed sets \(F_k\subset E_k\), such that \(m(E_k\backslash F_k)<\delta/2^{k+1}\), and \(f\in C(F_k)\). Consider define \(E_\delta=\bigcup_{k=1}^\infty F_k\) (see difference with (ii)) which is just a measurable set, or to be more specific, a \(F_\sigma\) set. But again by using Theorem of Approximation for measurable sets, there exists a closed set \(F\subset E_\delta\), such that \(m(E_\delta-F)<\delta/2\).

Now we check the measure of the removal part. Since

\[ E\backslash F \subset (E\backslash E_\delta)\cup (E_\delta\backslash F), \]

we have

\[ \begin{align*} m(E\backslash F)&\leq m\left[E\backslash E_\delta\right]+m\left[E_\delta \backslash F\right]\\ &=m\left[E\backslash \left(\bigcup_{k=1}^\infty F_k\right)\right]+m\left[E_\delta \backslash F\right]\\ &\leq m\left[\bigcup_{k=1}^\infty (E_k\backslash F_k)\right] + \frac{\delta}{2}\\ &\leq \sum_{k=1}^\infty m(E_k\backslash F_k)+\frac{\delta}{2}\\ &=\frac{\delta}{2}+\frac{\delta}{2}=\delta \end{align*} \]

Thus we are done.

Continuous Extension Theorem

Assume \(f(x)\) is a measurable function bounded \(a.e.\) on measurable set \(E\subset \mathbb{R}\), then \(\forall \delta>0\), there \(\exists\) closed set \(F_\delta\subset E\) and continuous function \(g(x)\) on whole space \(\mathbb{R}\), such that \(m(E-F_\delta)<\delta\) and \(g(x)=F(x),\forall x\in F_\delta\), and

\[ \begin{equation} \sup_{x\in \mathbb{R}}g(x)=\sup_{x\in F_\delta}f(x),\quad \inf_{x\in \mathbb{R}}g(x)=\inf_{x\in F_\delta}f(x).\label{boundary-condition} \end{equation} \]

which naturally satisfies the condition in the theorem.

Proof

By Lusin Theorem, \(\forall \delta>0\), there exists closed set \(F\subset E\), such that \(m(E-F)<\delta\), and \(f\in C(F)\). Now the only problem is, how to extend this continuoud function on \(F\) to all space \(F\) and satisfy the so-called boundary condition \(\ref{boundary-condition}\).

By Conposition of closed set, \(F\) could be expressed by \(\mathbb{R}-G\), where \(G\) is a union of denumerable open intervals which are mutually disjoint. That is,

\[ F^c=\mathbb{R}-F=\bigcup_{i=1}^\infty(a_i,b_i). \]

where at most \(2\) open intervals are infinite intervals. From here we formulate a function on \(\mathbb{R}\)

\[ g(x)=\begin{cases} f(x),\quad &x\in F,\\ f(a_i)+\frac{f(b_i)-f(a_i)}{b_i-a_i}(x-a_i),\quad & x\in (a_i,b_i),\\ f(a_i),\quad &x\in (a_i,+\infty),\\ f(b_i),\quad &x\in (-\infty, b_i). \end{cases} \]

Convergence in Measure

Example. \(\forall n\in \mathbb{N}^+\), \(\exists ! k,i\in \mathbb{N}^+\), such that \(n=2^k+i\), where \(0\leq u\leq 2^k\). Formulate a function sequence

\[ f_n(x)=\chi_{\left[\frac{i}{2^k},\frac{i+1}{2^k}\right]}(x),\quad n=1,2,\cdots,\quad x\in [0,1] \]

So for every point \(x\in [0,1]\), \(f_n(x)\) does not converge.

For the above example, we could consider its convergence in measure, or probability.

Still the above example.

If we define a set

\[ E_n(\varepsilon)=\{x\in [0,1]: |f_n(x)-0|\geq \varepsilon\}, \]

then its measure

\[ m(E_n(\varepsilon))=\frac{1}{2^k}\rightarrow 0 \]

which means for sufficient large \(n\), the frequency of \(0\) is close to \(1\). So we call \(\{f_n(x)\}\) converges to \(f(x)=0\) in measure.

Definition of convergence in measure

Assume function \(f\) and function sequence \(\{f_n\}_{n\geq 1}\) are measurable function bounded \(a.e\) on \(D\). If \(\forall \delta>0\), we have

\[ \lim_{n\rightarrow \infty} m(\{|f_n-f|\geq \delta\})=0 \]

then we call \(\{f_n\}\) converges to \(f\) in measure on region \(D\).

Using \(\varepsilon-N\) language, we have

\[ \forall \varepsilon>0, \sigma>0,\exists N\in \mathbb{N}^+, \forall n\geq N, m(\{|f_n-f|>\sigma\})<\varepsilon. \]

Generally speaking, "convergence point-wisely" has no apparent relationship with "convergence in measure", readers could see the following exmaple.

Example. Define \(f_n(x)=\chi_{(n,\infty)}(x)\), \(f(x)=0\), then \(\forall x\in \mathbb{R}\), we have \(f_n(x)\rightarrow f(x)\). But if we let \(\delta=1/2\), then \(\{|f_n-f|\geq 1/2\}=(n,\infty)\), whose measure \(m(\{|f_n-f|\geq 1/2\})=\infty\), which does not converges to \(f(x)\) in measure.

Uniqueness of Convergent Function

Uniqueness of Convergent Function

If \(f_n(x)\) converges to \(f(x)\) and \(g(n)\) on \(E\) in measure, then

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

Proof

Because \(|f(x)-g(x)|\leq |f(x)-f_n(x)|+|f_n(x)-g(x)|\) (at least one set of the right side should be larger then \(\varepsilon/2\) if \(|f(x)-g(x)|>\varepsilon\)), so

\[ \{|f(x)-g(x)|>\varepsilon\}\subset \left\{|f(x)-f_n(x)|>\frac{\varepsilon}{2}\right\}\cup \left\{|g(x)-f_n(x)|>\frac{\varepsilon}{2}\right\} \]

So by subadditivity, we have

\[ m(\{|f(x)-g(x)|>\varepsilon\})\leq m\left(\left\{|f(x)-f_n(x)|>\frac{\varepsilon}{2}\right\}\right)+m\left(\left\{|g(x)-f_n(x)|>\frac{\varepsilon}{2}\right\}\right)\rightarrow 0,\quad n\rightarrow \infty \]

Relationship between two Convergences

Lebesgue's Theorem: Relationship between two Convergences

Assume \(m(E)<\infty\), \(\{f_n(x)\}\) is a measurable function sequence \(a.e.\) on \(E\). If \(\{f_n\}\) converges to function \(f(x)\) \(a.e.\) on \(E\), where \(f(x)\) is a bounded function \(a.e.x\in E\), then \(f_n(x)\) converges to \(f(x)\) in measure.

HintsProof

Use \(m(E)<\infty\) to show that \(f_n(x) \rightrightarrows f(x)\), which could gives convergence in measure.

By Egoroff Theorem, because \(m(E)<\infty\), we have point-wise convergent function sequence \(\{f_n\}\) also converges uniformly to \(f(x)\). That is, we have \(\forall \delta>0\), \(\exists\) measurable set \(E_\delta\), such that \(m(E_\delta)<\delta\) and

\[ f_n(x)\rightrightarrows f(x),\quad \forall x\in E\backslash E_\delta. \]

Then by definition of uniform convergence, we have \(\forall \varepsilon>0\), \(\exists N>0\), \(\forall n\geq N\),

\[ |f_n(x)-f(x)|<\varepsilon,\quad x\in E\backslash E_\delta \]

which means the set

\[ \{x\in E: |f_n(x)-f(x)|\geq \varepsilon\}\subset E_\delta \]

so

\[ m(\{x\in E: |f_n(x)-f(x)|\geq \varepsilon\})\leq m(E_\delta)<\delta. \]

and we are done.

Note that \(m(E)<\infty\) is not necessary, see that we use the uniform convergence to prove, which is the most important.

Riesz Theorem

Assume function \(f\) and function sequence \(\{f_n\}\) are measurable function \(a.e.\) on \(E\), where \(f\) and \(\{f_n\}\) are bounded \(a.e.\) on \(E\). Then

(i) If \(\{f_n\}\) converge to \(f\) in measure, then there exists subsequence \(\{f_{n_j}\}\) convergent to \(f\) \(a.e.x\in E\).

(ii) If \(m(E)<\infty\), and \(\{f_n\}\) converge to \(f\) \(a.e.x\in E\), then \(\{f_n\}\) converge to \(f\) in measure.

Proof for (i)Proof for (ii)

We hope to use \(\delta\) to get a subsequence of \(f_n\).

\(\forall k\in \mathbb{N}^+\), let \(\delta_k=\frac{1}{2^k}\), so \(f_n\overset{m}{\rightarrow}f\) implies

\[ \lim_{n\rightarrow \infty}m(\{|f_n-f|\geq \delta_k\})=0. \]

So there exsits \(n_k\), such that

\[ m(\{|f_{n_k}-f|\geq \delta_k\})<\frac{1}{2^k}. \]

Notice {m({|f_n-f|\geq \delta_k})}_{n\geq 1} is a number sequence, so \(n_1<n_2<\cdots\).

Define

\[ E:=\bigcap_{p=1}^\infty\bigcup_{k=p}^\infty\left\{|f_{n_k}-f|\geq \frac{1}{2^k}\right\} \]

and for every \(p\), measure

\[ m(E)\leq m\left(\left\{|f_{n_k}-f|\geq \frac{1}{2^k}\right\}\right)<\sum_{k=p}^\infty\frac{1}{2^k}=\frac{1}{2^{p-1}}. \]

let \(p\rightarrow \infty\), we have \(m(E)=0\). Thus for every \(x\in D-E\), i.e.

\[ x\in \bigcup_{p=1}^\infty\bigcap_{k=p}^\infty\left\{|f_{n_k}-f|< \frac{1}{2^k}\right\} \]

which means, \(\exists p_0\geq 1\), such that

\[ |f_{n_k}-f|< \frac{1}{2^k},\quad \forall k\geq p_0 \]

meaning \(\lim\limits_{k\rightarrow \infty}f_{n_k}=f\).

\(\forall \delta>0\), \(\forall \varepsilon>0\), since \(m(D)<\infty\), by Egoroff Theorem, there exists measurable set \(E\subset D\), such that \(m(D-E)<\varepsilon\), and \(f_n\rightrightarrows f, a.e.x\in E\). That is, there exsits \(N\), such that

\[ |f_n(x)-f(x)|<\delta,\quad x\in E,n\geq N. \]

which means

\[ m(\{|f_n(x)-f(x)|\geq \delta\})=M(D-E)<\varepsilon,\quad n>N \]

meaning

\[ \lim_{n\rightarrow \infty}m(\{|f_n(x)-f(x)|\geq \delta\})=0. \]

Lemma: Sufficient and Necessary Condition for convergence in measure

Assume function \(f\) and function sequence \(\{f_n\}\) are measurable on \(E\) where \(m(E)<\infty\). Then \(f_n(x)\) converges to \(f(x)\) in measure, iff \(\forall\) subsequence \(\{f_{n_j}\}\), \(\exists\) a subsequence \(\{f_{n_{j_i}}\}\) such that

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

Proof

• "\(\Rightarrow\)".

We prove \(\forall\) subsequence \(\{f_{n_j}\}\), \(f_{n_j}\) converges to \(f\) in measure, which holds apparently because \(f_n(x)\) converges to \(f(x)\) in measure.

Then apply Riesz Theorem to \(f_{n_j}\) and we are done.

• "\(\Leftarrow\)". We make use of contradiction. That is, if \(\{f_n\}\) does not converges to \(f\) in measure, i.e. \(\exists \varepsilon_0>0\),

\[ \lim_{n\rightarrow \infty} m(\{|f_n-f|\geq \varepsilon_0\})\neq 0. \]

So \(\forall j, \exists n_j\), s.t.

\[ \begin{align} m(\{|f_{n_j}-f|\geq \varepsilon_0\})>0.\label{contradiction} \end{align} \]

But if there exists a subsequence \(\{f_{n_{j_i}}\}\) such that

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

then by Lebesgue's Theorem (\(m(E)<\infty\)), we have \(f_{n_{j_i}}\overset{m}{\rightarrow}f\), which contradicts inequation \(\ref{contradiction}\).