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 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

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