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Differential & Integral

From we already known, differential operation and integration operation could be inverse in terms of Riemann integral.

Riemann Integration

(i) If \(f\in R[a,b]\), and is continuous at \(x_0\), then

\[ F(x)=\int_a^x f(t)dt \]

is differentiable at \(x_0\), and \(F'(x_0)=f(x_0)\).

(ii) If \(f\in C^1[a,b]\), and \(f'\in R[a,b]\), then \(\forall x\in [a,b]\), we have

\[ \int_a^xf'(t)dt=f(x)-f(a). \]

In this chapter, we aim to extend the above theorem in terms of Lebesgue Integration.

Vitali's Covering Theorem

Vitali Covering

Assume \(E\subset \mathbb{R}\), \(\Lambda=\{I_\alpha\}\) is a family of intervals. If \(\forall x\in E\), \(\forall \varepsilon>0\), \(\exists I_\alpha\in \Lambda\), such that \(x\in I_\alpha\), \(m(I_\alpha)<\varepsilon\), then we call \(\Lambda\) is a covering in a sense of Vitali.

The following theorem is the main result in a sense of Vitali.

Vitali's Covering Theorem

Assume \(E\subset \mathbb{R}\), and \(m^*(E)<+\infty\). If \(\Lambda\) is a Vitali covering of \(E\), then \(\forall \varepsilon>0\), there exist finite number of intervals \(\{I_k\}_{1\leq k\leq n}\in \Lambda\) which are mutually disjoint, such that

\[ m\left(E-\bigcup_{k=1}^n I_k\right)<\varepsilon. \]

Dini Differential Quotient

Dini Differential Quotient

Assume \(f\) is defined on a neighborhood of \(x_0\in \mathbb{R}\), define

(i) upper right DQ: \(D^+f(x_0)=\overline{\lim}\limits_{h\rightarrow 0^+}\frac{f(x_0+h)-f(x_0)}{h}\),

(ii) lower right DQ: \(D_+f(x_0)=\underline{\lim}\limits_{h\rightarrow 0^+}\frac{f(x_0+h)-f(x_0)}{h}\),

(iii) upper left DQ: \(D^-f(x_0)=\overline{\lim}\limits_{h\rightarrow 0^-}\frac{f(x_0+h)-f(x_0)}{h}\),

(iv) lower left DQ: \(D_-f(x_0)=\underline{\lim}\limits_{h\rightarrow 0^-}\frac{f(x_0+h)-f(x_0)}{h}\).

There are some properties.

Properties of Dini Differential Quotient

(i) \(D^+f(x_0)\geq D_+f(x_0)\), \(D^-f(x_0)\geq D_-f(x_0)\).

(ii) \(D^+(-f)=-D_+f\), \(D^-(-f)=-D^-f\).

Example. Calculate Dini Differential Quotient of

\[ f(x)=\begin{cases} \sin \frac{1}{x},\quad &x\neq 0,\\ 0,\quad &x=0. \end{cases} \]

and \(g(x)=xf(x)\).

\(D^+f=D^-f=+\infty\), \(D_+f=D_-f=-\infty\).

\(D^+g=D^-g=+1\), \(D_+g=D_-g=-1\).

Monotonic Real Function

For monotonically increasing function, we could prove it has differential quotient.

Lemma: Monotonically increasing function has Dini DQ

Assume \(f\) is a monotonically increasing real function on \([a,b]\), then \(f'(x)\) exists \(a.e.\).

For monotonically increasing function \(f\), to show that its four Dini differential quotients are of the same, we only have to show that

\[ D^+f\leq D_-f,\quad D^-f\leq D_+f. a.e.x\in [a,b]. \]

because the above two inequation combined with (i) in Property of Dini differential quotient could deduce "=".

In a nutshell, we have to show that

\[ m(E_1)=m(E_2)=0, \]

where

\[ E_1=\{x: D^+f> D_-f\}, E_2=\{D^-f> D_+f\}. \]

We prove this using Vitali's covering Theorem.

Finally, we have the following theorem.

Lebesgue Monotonic Differential Theorem

Assume \(f(x)\) is a monotonically increasing real function on \([a,b]\), then \(f'(x)\) exists \(a.e.\), and \(f'\in L[a,b]\), and

\[ \int_{[a,b]}f'(x)dx\leq f(b)-f(a). \]

Let

\[ f_n(x)=n\left[f\left(x+\frac{1}{n}\right)-f(x)\right], \quad x\in [a,b]. \]

and \(f(x)=f(b), x>b\). So because \(f\) is monotonically increasing, \(f_n(x)\leq 0\), and from Lemma which tells us \(f\) has differential quotient, we have

\[ \lim_{n\rightarrow \infty}f_n(x)=f'(x), \quad a.e.x\in [a,b]. \]

By Fatou Lemma, we have

\[ \begin{align*} \int_a^b f'(x)dx&\leq \underline{\lim}\limits_{n\rightarrow \infty}\int_a^b f_n(x)dx\\ &=\underline{\lim}\limits_{n\rightarrow \infty}\int_a^b n\left[f\left(x+\frac{1}{n}\right)-f(x)\right] dx\\ &=\underline{\lim}\limits_{n\rightarrow \infty}\left[ n\int_b^{b+\frac{1}{n}}f(x)dx-n\int_a^{a+\frac{1}{n}}f(x) dx\right]\\ &=\underline{\lim}\limits_{n\rightarrow \infty}\left[ f(b)- n\int_a^{a+\frac{1}{n}}f(x) dx\right]\quad\text{by }f(x)=f(b), x>b\\ &\leq f(b)-f(a). \end{align*} \]

See the following typical examples to check the condition and conclusion of the above theorem.

Example. Assume \(E\subset [a,b]\) is a zero-measure set. Then there exists a monotonically increasing real continuous function \(f\) on \([a,b]\), such that for every \(x\in E\), we have \(f'(x)=\infty\).

For zero-measure set \(E\), there exists a monotonically decreasing sequence of sets \(\{G_n\}_{n\geq 1}\), such that

\[ E\subset G_n,\quad G_n<\frac{1}{2^n}. \]

If we let

\[ f_n(x)=m([a,x]\cap G_n), \quad f(x)=\sum_{i=1}^\infty f_n(x). \]

Similar to proof in Example for measure of interval intersection, we could prove that \(f_n\) is non-negative monotonically increasing continuous function which is not generalized(convergent, so we could change order in series).

Now for all \(x\in E\), \(n\geq 1\), so long as \(h\) is sufficiently small, such that \([x,x+h]\subset G_n\). Because \(G_k\) is monotonically decreasing, so \([x,x+h]\subset G_k\), for \(k=1,2,\cdots,n\). So differential quotient

\[ \begin{align*} \frac{f(x+h)-f(x)}{h}&=\frac{1}{h}\left[\sum_{k=1}^\infty f_k(x+h)- \sum_{k=1}^\infty f_k(x)\right]\\ &=\frac{1}{h}\sum_{k=1}^\infty \left[ f_k(x+h)-f_k(x)\right]\\ &\geq \frac{1}{h}\sum_{k=1}^n \left[ f_k(x+h)-f_k(x)\right]\\ &=\frac{1}{h}\sum_{k=1}^n m([x,x+h]\cap G_k)\\ &=\frac{1}{h}\sum_{k=1}^n m([x,x+h])\\ &=\frac{nh}{h}=n \end{align*} \]

So \(f'(x)=\lim\limits_{n\rightarrow \infty}n=\infty\).

This example shows that in Lebesgue Monotonic Differential Theorem, we could not weaken condition "\(f\) is differential \(a.e.x\in E\)" any more.

Example. Prove: There exists a strictly monotonically increasing continuous function on \([a,b]\), such that

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

This example shows that we could not change "\(\leq\)" into "\(=\)" in conclusion of Lebesgue Monotonic Differential Theorem

Bounded Variation Function

Assume \(f(x)\) is a real function (not genaralized) on \([a,b]\), we make a partition \(\Delta: a=x_0<x_1<\cdots<x_n=b\) and sum \(|f(x_i)-f(x_{i-1})|\) up, denoted by

\[ V_\Delta(f)=\sum_{i=1}^\infty|f(x_i)-f(x_{i-1})|, \]

which is called variation of \(f(x)\) on \([a,b]\) corresponding to partition \(\Delta\). Apparently, a variation of a function is dependent on its partition \(\Delta\), so we want to get rid of its influence, by defining its Total Variation as

\[ V_a^b(f):=\sup\{V_\Delta: \forall \Delta \text{ of } [a,b]\}. \]

If \(V_a^b<\infty\), then we call \(f(x)\) is a bounded variation function. The whole set of bounded variation functions on \([a,b]\) is denoted by \(BV([a,b])\).

Corollary: Monotonic real function must be bounded variation function

Monotonic real function (not generalized) must be bounded variation.

\(V_\Delta(f)=|f(b)-f(a)|<\infty\).

While monotonic real function is bounded variation function, it is not so for continuous function, as shown in the following example.

Example. Assume

\[ f(x)=\begin{cases} x\cos\frac{\pi}{2x},\quad &x\in(0,1],\\ 0,\quad &x=0. \end{cases} \]

Then Prove: \(f(x)\) is continuous on \([0,1]\), but is not bounded variation.

For \(n\geq 1\), construct a partition

\[ \Delta_n:\left\{0,\frac{1}{2n},\frac{1}{2n-1},\cdots,\frac{1}{2},1 \right\} \]

then

\[ \begin{align*} V_\Delta (f)&=\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2(n-1)}+\frac{1}{2(n-1)}+\cdots+\frac{1}{2}+\frac{1}{2}\\ &=\sum_{i=1}^n \frac{1}{i}\rightarrow \infty (n\rightarrow \infty) \end{align*} \]

So it is not bounded variation.

Properties for bounded variation function

(i) Bounded variation function on \([a,b]\) must be bounded.

(ii) Operations. If \(f\) and \(g\) are both bounded variation function, then \(f\pm g\) and \(fg\) both are bounded variation functions. And if \(|g(x)|\geq \lambda>0\), then \(f/g\) are also bounded variation function.

(iii) Region additivity. If \(f\) is a function on \([a,b]\), then \(\forall c\in [a,b]\),

\[ V_a^b(f)=V^c_a(f)+V^b_c(f). \]

(i) \(\forall a\in [a,b]\), we have

\[ |f(x)-f(a)|<V_a^b (f)<\infty \]

so \(|f(x)|\leq |f(a)|+V_a^b (f)\).

(ii) We could use

\[ \begin{align*} V_\Delta(f+g)&=\sum_{i}|(f+g)(x_i)-(f+g)(x_{i-1})|\\ &\leq \sum_{i}|f(x_i)-f(x_{i-1})| + \sum_{i}|g(x_i)-g(x_{i-1})|\\ &=V_\Delta(f) + V_\Delta(g)\leq V_a^b(f)+V^b_a(g) \end{align*} \]

to show that \(f+g\) is also bounded variation function.

(iii) For "\(\leq\)", we use \(\tilde{\Delta}=\Delta\cup \{c\}\).

For "\(\geq\)", we use \(\Delta=\Delta_1\cup \Delta_2\).

From property (iii), we could know that \(\forall x\in [a,b]\), \(f\) is also bounded variation on \([a,x]\). So function \(F(x)=V_a^x(f)\) is called incomplete bounded variation function, which is apparently a non-negative function and monotonically increases.

Continuity relation between \(f\) and \(V_a^x(f)\)

If \(f\) is a bounded variation function on \([a,b]\), then \(f(x)\) and \(V_a^x(f)\) have same left and right continuous points.

  • Points from \(V_a^x(f)\) to \(f\).

Assume \(x_0\in [a,b)\) is a right continuous point of \(V_a^x(f)\), then

\[ |f(x_0+h)-f(x_0)|\leq V_{x_0}^{x_0+h}(f)=V_{a}^{x_0+h}(f)-V_{a}^{x_0}(f). \]

So \(x_0\) is also the right continuous point of \(f\). Similar logic for left continuous points.

  • Points from \(f\) to \(V_a^x(f)\).

Assume \(x_0\in[a,b)\) is a right continuous point of \(f\). Then \(\forall \varepsilon>0\), \(\exists \delta>0\), such that

\[ \begin{align} |f(x)-f(x_0)|<\frac{\varepsilon}{2},\quad x_0<x<x_0+\delta. \label{right-continuous-point} \end{align} \]

and because \(\lim_{x\rightarrow x_0^+}V_a^x(f)<\infty\)

\[ \begin{align} 0<V_a^{x_2}(f)-V_a^{x_1}(f)<\frac{\varepsilon}{2},\quad x_0<x_1<x_2<x_0+\delta. \label{F-continuous} \end{align} \]

\(\forall x\in (x_0,x_0+\delta)\), for a parition \(\Delta: \{y_k\}_{0\leq k\leq n}\)of \([x_0,x]\):

\[ x_0=y_0<y_1<\cdots <y_n=x, \]

by inequation \(\ref{right-continuous-point}\), we have

\[ |f(y_1)-f(y_0)|<\frac{\varepsilon}{2}. \]

by inequation \(\ref{F-continuous}\), we have

\[ \sum_{k=2}^n |f(y_k)-f(y_{k-1})|\leq V_{y_1}^x(f)=V_a^{x}(f)-V_a^{y_1}(f)<\frac{\varepsilon}{2}. \]

so combine the above two inequation, we have

\[ \sum_{k=1}^n|f(y_k)-f(y_{k-1})|<\varepsilon. \]

By the arbitrariness of partition, we have

\[ V_a^x(f)-V_a^{x_0}(f)=V_{x_0}^x(f)\leq \varepsilon, \quad x_0<x<x_0+\delta. \]

So \(V_a^x(f)\) is right continuous at \(x_0\). Similar logic for left continuous point. We are done.

From the above theorem, we could have the following corollary.

Corollary: bounded variation function has at most denumerable discontinuity points

If \(f\) is bounded variation on \([a,b]\), then the number of discontinuity points of \(f\) is at most denumerable.

\(f\) has the same discontinuity points with \(V_a^x(f)\), while the latter is a monotonically increasing function, which has at most denumerable discontinuity points as Example shows.

Jordan Decomposition Theorem: description of bounded variation function

\(f\) is a bounded variation function, iff \(f\) could be expressed by the subtraction of two monotonically increasing real functions.

  • "\(\Leftarrow\)". \(g,h\) are monotonically increasing, so \(g, h\) are both bounded variation functions. By property (ii), we could know that \(f=g-h\) is also bounded variation function.

  • "\(\Rightarrow\)". Assume \(f\) is bounded variation, then \(\forall a\leq x_1\leq x_2\leq b\),

\[ f(x_2)-f(x_1)\leq V_{x_1}^{x_2}(f)=V_a^{x_2}(f)-V_a^{x_1}(f), \]

which means \(V_a^{x_1}(f)-f(x_1)\leq V_a^{x_2}(f)-f(x_2)\).

so define \(g=V_a^x(f)\), \(h=V_a^x(f)-f(x)\), both are monotonically increasing. Then \(f=g-h\) and we are done.

Corollary: bounded variation function must be differential a.e.

Bounded variation function \(f\) on \([a,b]\) must be differential a.e. and its derivative \(f'\) is L integrable.

\(f=g-h\), where \(g,h\) are monotonically increasing function. By Lebesgue Monotonic Differential Theorem, \(g,h\) is differential \(a.e.\) on \([a,b]\), and their derivatives \(g',h'\) are also L integrable. So by operations, \(f\) is also differential \(a.e.\) on \([a.b]\) and \(f'=f'-h'\) is also L integrable.

Indefinite Integration & its Differential

Definition of Indefinite Integration

If \(f\) is L integrable on \([a,b]\), then

\[ F(x)=\int_a^x f(t)dt, \quad x\in [a,b] \]

is called Indefinite Integration of \(f\).

Using what we have known from above, we have the following property for Indefinite integration.

Property of Indefinite Integration

If \(f\) is L integrable on \([a,b]\), then its indefinite integration \(F(x)\) is a continuous bounded variation function on \([a,b]\) and satisfies

\[ V_a^b(f)=\int_a^b|f(t)|dt. \]

\(F\) is continuous because of absolute continuity of L integral.

To get to the result of this part, we should have a lemma proved.

Lemma: indefinite integration equals zero

Assume \(f\) is L integral on \([a,b]\), and its indefinite integration

\[ F(x)\equiv 0, \]

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

For zero-constant function \(F\), we have its total variation function \(V_a^b(F)=0\). By Property of Indefinite Integration, we have

\[ \int_a^b|f(t)|dt=0. \]

Then by Example in Non-negative measurable function, we have

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

and we are done.

Finally we have the following theorem.

Theorem for Derivative of indefinite integration

Assume \(f\) is L integrable on \([a,b]\), and \(F(x)\) is an indefinite integration of \(f\), then

\[ F'(x)=f(x),\quad a.e.x\in [a,b]. \]

Since \(F\) is bounded variation and continuous on \([a,b]\), then by Corollary of Jordan Decomposition, we have \(F\) has differential \(a.e.x\in [a,b]\) and its derivative \(F'\) is L integrable.

  • \(f\) is bounded. So \(|f(x)|\leq M\).

Let

\[ g_n(x)=n\left[F\left(x+\frac{1}{n}\right)-F(x)\right], \]

then

\[ \lim_{n\rightarrow \infty}g_n(x)=F'(x),\quad a.e.x\in [a,b]. \]

We check \(g_n\) for its condition of LDCT. That is,

\[ |g_n(x)|=\left|n\left[F\left(x+\frac{1}{n}\right)-F(x)\right]\right|=\left|n\int_x^{x+\frac{1}{n}}f(t)dt\right|\leq M. \]

So by LDCT, we have

\[ \begin{align*} \int_a^x F'(t)dt&=\lim_{n\rightarrow \infty}\int_a^xg_n(t)dt\\ &=\lim_{n\rightarrow \infty}n\int_a^x\left[F\left(t+\frac{1}{n}\right)-F(t)\right]dt\\ &=\lim_{n\rightarrow \infty}\left[n\int_a^x F\left(t+\frac{1}{n}\right)dt-n\int_a^x F(t)dt\right]\\ &=F(x)-F(a)=\int_a^xf(t)dt. \end{align*} \]
  • \(f\) is measurable. Let \(f\geq 0\) might as well.

Absolute Continuous Function

Definition of Absolute continuous function

Assume \(f\) is a real function on \([a,b]\). If \(\forall \varepsilon>0\), \(\exists \delta>0\), such that for finite number of open intervals \(\{(a_k,b_k)\}_{1\leq k\leq n}\) which are mutually disjoint, so long as \(\sum_{k=1}^n |b_k-a_k|<\delta\),

\[ \sum_{k=1}^n|f(b_k)-f(a_k)|<\varepsilon, \]

then we call \(f\) is an absolute continuous function.

Corollary: Indefinite integration of L-integrable function must be absolute continuous

Assume \(f\) is L integrable on \([a,b]\), then its indefinite integration \(F\) is absolute continuous.

Notice that

\[ \begin{align*} \sum_{k=1}^n|F(b_k)-F(a_{k})|&=\sum_{k=1}^n\left|\int_{a_{k}}^{b_k} f(t)dt\right|\\ &\leq \sum_{k=1}^n\int_{a_{k}}^{b_k} |f(t)|dt\\ &=\int_{\bigcup\limits_{k=1}^n (a_k,b_k)} |f(t)|dt \end{align*} \]

By absolute continuity of L integral, \(\forall \varepsilon>0\), \(\exists \delta>0\), such that \(\forall \{a_k,b_k\}\) satisfying \(\sum_{k=1}^n(b_k-a_k)<\delta\), we have

\[ \sum_{k=1}^n|F(b_k)-F(a_{k})|\leq \int_{\bigcup\limits_{k=1}^n (a_k,b_k)} |f(t)|dt<\varepsilon. \]

For absolute continuous function, we have the following properties.

Properties of Absolute continuous function

(i) Absolute continuous function must be continuous and consistently continuous.

(ii) Operations. If \(f\) and \(g\) are absolute continuous functions, then \(f\pm g\), \(fg\) are absolute continuous. And if \(g\) has no zero points, \(f/g\) is also absolute continuous.

(iii) Function composition. If \(f\) and \(\varphi\) are absolute continuous on \([a,b]\) and \([p,q]\) respectively, and \(a\leq \varphi(x)\leq b\), \(\varphi\) is strictly monotonically increasing, then function

\[ f(\varphi(t)) \]

is absolute continuous on \([p,q]\).

(iv) If \(f\) is absolute continuous function on \([a,b]\), then it must be bounded variation function.

(i) Choose \(n=1\) in definition.

(ii) Use some formula of triangle inequation.

(iii)

(iv) .

Mean Value Theorem for Integration

First mean value theorem for integration

Assume \(f\) is continuous on \([a,b]\), \(g\) is non-negative and L integrable on \([a,b]\). Then \(\exists \xi\), such that

\[ \int_a^bf(x)g(x)dx=f(\xi)\int_a^b g(x)dx. \]

  • \(\int_a^b g(x)dx=0\). It holds apparently.

  • \(\int_a^b g(x)dx\neq 0\).

Second mean value theorem for integration

Assume \(f\) is L integrable, \(g\) is monotonic, then \(\exists \xi\), such that

\[ \int_a^bf(x)g(x)dx=g(a)\int_a^\xi f(x)dx+g(b)\int_\xi^bf(x)dx. \]

Let \(g\) is monotonically increasing.

  • (i) \(g\in AC[a,b]\). Use Integration by parts.

Then \(g'\geq 0\) and \(g'\in L[a,b]\). Since \(f\in L[a,b]\), we have \(F(x)=\int_a^xf(t)dt\in AC[a,b]\) and \(F'(x)=f(x),\quad a.e.x\in [a,b]\).

So

\[ \begin{align*} \int_a^bf(x)g(x)dx&=\int_a^bF'(x)g(x)dx\\ &=F(x)g(x)|_a^b-\int_a^bF(x)g'(x)dx\\ &=F(b)g(b)-F(\xi)\int_a^bg'(x)dx\\ &=F(b)g(b)-F(\xi)(g(b)-g(a)) \end{align*} \]
  • (ii) \(g\) is just monotonically increasing.

Use a partition \(\Delta: a=x_0<x_1<\cdots<x_n=b\), where \(x_k=a+k(b-a)/n, k=1,2,\cdots,n\).

Let

\[ g_n(x)=\begin{cases} g(x),\quad &x=x_k,0\leq k\leq n,\\ g(x_{k-1})+\frac{n(g(x_k)-g(x_{k-1}))}{(b-a)}(x-x_{k-1}),\quad &x\in (x_{k-1},x+k). \end{cases} \]

Check the convergence of \(g_n\). If \(x\) is the continuity points, then \(\lim\limits_{n\rightarrow \infty}g_n(x)=g(x)\). If \(x\) is discontinuity point, it is at most denumerable, which means it is zero-measure set, meaning that

\[ \lim\limits_{n\rightarrow \infty}g_n(x)=g(x),\quad a.e.x\in [a,b]. \]

We know that \(g_n\in AC[a,b]\), so by (i)

\[ \int_a^bf(x)g_n(x)dx=g_n(a)\int_a^{\xi_n} f(x)dx+g_n(b)\int_{\xi_n}^bf(x)dx. \]

We want to use LDCT. Since \(f(x)g_n(x)\leq |f(x)|\max\{g(b),g(a)\}\in L[a,b]\),

\[ \begin{align*} \int_a^bf(x)g(x)dx&=\lim_{n\rightarrow\infty}\int_a^bf(x)g_n(x)dx\\ &=\lim_{n\rightarrow\infty}\left[g_n(a)\int_a^{\xi_n} f(x)dx+g_n(b)\int_{\xi_n}^bf(x)dx\right]\\ &=g(a)\int_a^\xi f(x)dx+g(b)\int_\xi^bf(x)dx. \end{align*} \]