Lecture 11

Recap

I(x)=\begin{cases} 0 & x\leq 0 \\ 1 & x>0 \end{cases}

Continue on Chapter 6

The step function

Theorem 6.16

If $\sum c_n$ converges and $\{s_n\}$ is a sequence of distinct elements of $(a,b)$ , and $f$ is continuous on $[a,b]$ , and $\alpha(x)=\sum_{n=1}^{\infty}c_nI(x-s_n)$ , then $\int_a^bf \ d\alpha=\sum_{n=1}^{\infty}c_nf(s_n)$ .

Proof:

For each $x$ , $I(x-s_n)\leq 1$ so $\sum_{n=1}^{\infty}c_nI(x-s_n)\leq \sum_{n=1}^{\infty}c_n$ converges (by comparison test).

Let $\epsilon>0$ . We can find $N$ such that $\sum_{n=N+1}^{\infty}c_n<\epsilon$ . (Recall that the series $\sum_{n=1}^{\infty}c_n$ converges if and only if $\lim_{N\to\infty}\sum_{n=1}^{N}c_n$ exists.)

Set $\alpha_1(x)=\sum_{n=1}^{N}c_nI(x-s_n)$ , and $\alpha_2(x)=\sum_{n=N+1}^{\infty}c_nI(x-s_n)$ .

Using the linearity of the integral, we have

\int_a^b f\ d\alpha_1= \sum_{n=1}^{N}c_n\int_a^b fd(I(x-s_n))= \sum_{n=1}^{N}c_nf(s_n)

On the other hand, with $M=\sup|f|$ ,

\left|\int_a^b f\ d\alpha_2\right|\leq \int_a^b |f|\ d\alpha_2\leq M\int_a^b \alpha_2\ dx=M\sum_{n=N+1}^{\infty}c_n(b-s_n)<\epsilon

So,

\begin{aligned} \left|\int_a^b f\ d\alpha-\sum_{n=1}^{\infty}c_nf(s_n)\right|&= \left|\int_a^b f\ d\alpha_2-\sum_{n=N+1}^{\infty}c_nf(s_n)\right|\\ &\leq |M\epsilon-\sum_{n=N+1}^{\infty}|c_n|M(b-s_n)|\\ &<2M\epsilon \end{aligned}

Since $\epsilon$ is arbitrary, we have $\int_a^b f\ d\alpha=\sum_{n=1}^{\infty}c_nf(s_n)$ .

Integration and differentiation

Theorem 6.20 Fundamental theorem of calculus

Let $f\in \mathscr{R}$ for $x\in [a,b]$ . We define $F(x)=\int_a^x f(t)\ dt$ . Then $F$ is continuous and if $f$ is continuous at $x_0\in [a,b]$ , then $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$ .

Proof:

Let $x<y,x,y\in [a,b]$ . Then,

|F(y)-F(x)|=\left|\int_a^y f(t)\ dt-\int_a^x f(t)\ dt\right|=\left|\int_x^y f(t)\ dt\right|\leq \sup|f|\cdot (y-x)

So, $F$ is continuous on $[a,b]$ .

Now, let $f$ be continuous at $x_0\in (a,b)$ and $\epsilon>0$ . Then we can find $\delta>0$ such that $a<x_0-\delta<s<x_0<t<x_0+\delta<b$ and $|f(u)-f(x_0)|<\epsilon$ for all $u\in (x_0-\delta,x_0+\delta)$ .

So,

\begin{aligned} \left|\frac{F(s)-F(x_0)}{s-x_0}-f(x_0)\right|&=\left|\frac{1}{s-x_0}\left(\int_a^s f(u)\ du-\int_a^{x_0} f(u)\ du\right)-f(x_0)\right|\\ &=\left|\frac{1}{s-x_0}\left(\int_s^{x_0} f(u)\ du\right)-\frac{1}{s-x_0}\left(\int_s^{x_0} f(x_0)\ dv\right)-f(x_0)\right|\\ &=\left|\frac{1}{x_0-s}\left(\int_s^{x_0} [f(u)-f(x_0)]\ du\right)\right|\\ &\leq \frac{1}{x_0-s}\epsilon(x_0-s)\\ &= \epsilon \end{aligned}

QED

If $f\in \mathscr{R}$ , and there exists a differentiable function $F:[a,b]\to \mathbb{R}$ such that $F'=f$ on $(a,b)$ , then

\int_a^b f(x)\ dx=F(b)-F(a)

Proof:

Let $\epsilon>0$ and $P=\{x_0,x_1,\cdots,x_n\}$ be a partition of $[a,b]$ .

By the mean value theorem, on each subinterval $[x_{i-1},x_i]$ , there exists $t_i\in (x_{i-1},x_i)$ such that

F(x_i)-F(x_{i-1})=F'(t_i)(x_i-x_{i-1})=f(t_i)\Delta x_i

Since $m_i\leq f(t_i)\leq M_i$ , notices that $\sum_{i=1}^n f(t_i)\Delta x_i=F(b)-F(a)$ .

So,

L(P,f)\leq F(b)-F(a)\leq U(P,f)

So, $f\in \mathscr{R}$ and $\int_a^b f(x)\ dx=F(b)-F(a)$ .

QED