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Math4121 Lecture 37

Extended fundamental theorem of calculus with Lebesgue integration

Density of continuous functions

Lemma for density of continuous functions

Let K\subseteq U be bounded sets in \mathbb{R}, K is closed and U is open. Then there is a continuous function g such that \chi_K\leq g\leq \chi_U.

Proof in homework.

Hint: Consider the basic intervals cases.

Theorem for continuous functions

Let f be integrable. For each \epsilon>0, there is a continuous function g:\mathbb{R}\to\mathbb{R} such that \int_{\mathbb{R}}|f-g|dm<\epsilon.

Proof:

First where f=\chi_S for some bounded means set S. then extended to all f integrable.

First, assume f=\chi_S. Let \epsilon>c, we can find K\subseteq S\subseteq U. and K is closed and U is open such that (by definition of Lebesgue outer measure)

m(K)+\frac{\epsilon}{2}>m(S)>m(U)-\frac{\epsilon}{2}

In particular, m(U\setminus K)=m(U)-m(K)<\epsilon.

By lemma, there is a continuous function g such that \chi_K\leq g\leq \chi_U.

So

\int_E |\chi_S -g|dm=\int_{U\setminus K} |\chi_S -g|dm\leq m(U\setminus K)<\epsilon

For the general case,

By the Monotone Convergence Theorem (use |f|\chi_{[-N,N]} to approximate |f|), we can find N large such that

\int_{E_N^c}|f|dm<\frac{\epsilon}{3}

where E_N=E\cap [-N,N].

Notice that by the definition of Lebesgue integral, \int f^+ dm=\sup\{\int \phi^+ dm:\phi\text{ is simple and } \phi\leq f^+\} and \int f^- dm=\sup\{\int \phi^- dm:\phi\text{ is simple and } \phi\leq f^-\}.

By considering f^+ and f^- separately, we can find a simple function \phi such that

\int_{E_N} |f-\phi|dm<\frac{\epsilon}{3}

For each i=1,2,\cdots,n, we can find g_i continuous such that

\int_{E}|\chi_{S_i}-g_i|dm<\frac{\epsilon}{3M}

where M=\sum_{i=1}^n |\alpha_i|.

Take g=\sum_{i=1}^n \alpha_i g_i,

\int_E |\phi-g|dm\leq \sum_{i=1}^n |\alpha_i|\int_E |g_i-\chi_{S_i}|dm<\frac{\epsilon}{3}

\phi-g=\sum_{i=1}^n \alpha_i (\chi_{S_i-g_i})

All in all,

\begin{aligned}
\int_E |f-g|dm&\leq \int_E|f-\phi|dm+\int_E |\phi-g|dm\\
&=\int_{E_N^c}|f|dm+\int_E |f-\phi|dm+\int_E |\phi-g|dm\\
&<\frac{\epsilon}{3}+\frac{\epsilon}{3}+\frac{\epsilon}{3}\\
&=\epsilon
\end{aligned}

QED

Road map for proving the fundamental theorem of calculus in Lebesgue integration

Recall the Riemann-Stieltjes integral:

If g\in \mathscr{R}(\alpha) on [a,b],

G(x)=\int_a^x g d\alpha,

then:

1. G is continuous on [a,b]
2. If g is continuous at x\in [a,b], then G is differentiable at x with G'(x)=g(x).

To extend this to the case where g is Lebesgue integrable, we use the Hardy-Littlewood maximal function.

Definition of the Hardy-Littlewood maximal function

Given an interval I\subseteq \mathbb{R}, define the averaging operator A_I f(x)=\frac{\chi_I(x)}{m(I)}\int_I f(x)dm.

(This function takes the average of f over the interval I.)

The Hardy-Littlewood maximal function is defined as:

f^*(x)=\sup_{I\text{ is open interval}}A_I f(x)

We will show that f^* is not that such worse than f. (Prove on Wednesday)

Relates to the Fundamental Theorem of Calculus in Lebesgue integration.

\frac{G(x+h)-G(x)}{h}=\frac{1}{h}\int_x^{x+h} g(t)dt=A_{[x,x+h]}g(x)

If we can control all the averages, we can control the function.