78 lines
2.7 KiB
Markdown
78 lines
2.7 KiB
Markdown
# Math4121 Lecture 39
|
|
|
|
## Fundamental theorem of calculus (In Lebesgue integration)
|
|
|
|
### Preliminary results
|
|
|
|
#### Lemma 1
|
|
|
|
Riemann integrable functions are Lebesgue integrable
|
|
$$\int_a^b f(x)dx = \int_{[a,b]} f dm$$
|
|
|
|
#### Lemma 2
|
|
|
|
Density of continuous functions: Given $f$ integrable, then $\exists \epsilon > 0$ there is $g$ continuous such that $\int_{[a,b]} |f-g| dm < \epsilon$
|
|
|
|
#### Lemma 3
|
|
|
|
Maximal function: $f^*(x) = \sup_{I\text{ is open intervals}}A_I f(x)$, where $A_I = \frac{\chi_I}{m(I)} \int_I f dm$. Then $|\{x\in\mathbb{R}:f^*(x)>\lambda\}|<\frac{2}{\lambda}\int_{\mathbb{R}}|f|dm$
|
|
|
|
#### Lemma 4
|
|
|
|
$I=[a,b]$, $I_\delta = [a+\delta, b-\delta]$, $\delta>0$, $\lim_{\delta\to 0^+} A_{I_\delta} f(x) = A_I f(x)$. (Prove via dominated convergence theorem)
|
|
|
|
> Riemann's Fundamental theorem of calculus:
|
|
>
|
|
> If $g$ is continuous on $[a,b]$, then $G(x) = \int_a^x g(t)dt$ is differentiable on $(a,b)$ and $G'(x) = g(x)$ for all $x\in(a,b)$.
|
|
|
|
### Lebesgue's Fundamental theorem of calculus
|
|
|
|
If $f$ is Lebesgue integrable on $[a,b]$, then $F(x) = \int_a^x f(t)dt$ is differentiable almost everywhere and $F'(x) = f(x)$ almost everywhere.
|
|
|
|
Outline:
|
|
|
|
Let $\lambda,\epsilon > 0$. Find $g$ continuous such that $\int_{\mathbb{R}}|f-g|dm < \frac{\lambda \epsilon}{5}$.
|
|
|
|
To control $A_I f(x)-f(x)=(A_I(f-g)(x))+(A_I g(x)-g(x))+(g(x)-f(x))$, we need to estimate the three terms separately.
|
|
|
|
Our goal is to show that $\lim_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I f(x)-f(x)|=0$. For $x$ almost every $x\in[a,b]$.
|
|
|
|
This implies the fundamental theorem of calculus.
|
|
|
|
Since $\frac{F(x+h)-F(x)}{h}=\frac{1}{m(I_h)}\int_{I_h}f dm$, if the above condition holds, then $\forall \eta>0$, we can find $r>0$ such that $\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I f(x)-f(x)|<\frac{\eta}{2}$.
|
|
|
|
Now given $h<\min\{r, x-a\}$, we can find by [4](https://notenextra.trance-0.com/Math4121_L39.html#Lemma-4) an interval $I_h^*$ such that
|
|
|
|
$$
|
|
\left|\frac{1}{h}\int_{I_h^*}f dm - \frac{1}{m(I_h^*)}\int_{I_h^*}f dm\right|<\frac{\eta}{2}
|
|
$$
|
|
|
|
Proof:
|
|
|
|
Let
|
|
|
|
$$
|
|
F=\left\{x\in [a,b]:\limsup_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I f(x)-f(x)|>\lambda \right\}
|
|
$$
|
|
|
|
Need to show $m(F)<\epsilon$.
|
|
|
|
Since $F\subseteq \{(f-g)^*>\frac{\lambda}{2}\}\cup \{(f-g)>\frac{\lambda}{2}\}$
|
|
|
|
$$
|
|
\limsup_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I f(x)-f(x)|\leq \limsup_{r\to 0^+}\sup_{I\text{ is open interval}, m(I)<r, x\in I}|A_I (f-g)(x)|+|g(x)-f(x)|
|
|
\leq |(f-g)^*(x)|+|(f-g)(x)|
|
|
$$
|
|
|
|
By maximal inequality and Markov's inequality,
|
|
|
|
$$
|
|
\begin{aligned}
|
|
m(F)&\leq \frac{4}{\lambda}\int_{\mathbb{R}}|f-g|dm+\frac{1}{\lambda}\int_{\mathbb{R}}|f-g|dm\\
|
|
&=\frac{5}{\lambda}\frac{\lambda \epsilon}{5}\\
|
|
&=\epsilon
|
|
\end{aligned}
|
|
$$
|
|
|
|
QED
|