3.1 KiB
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