Math4121 Lecture 35

Continue on Lebesgue Integration

Lebesgue Integration

Definition of Lebesgue Integral

For simple functions \phi = \sum_{i=1}^{n} a_i \chi_{S_i}, given a measure E, the Lebesgue integral is defined as:


\int_{\mathbb{R}^n} \phi \, dm = \sum_{i=1}^{n} a_i m(S_i\cap E)

Given a non-negative measurable function f and a measurable set E.

Define \int_E f \, dm = \sup \left\{ \int_E \phi \, dm : \phi \text{ is a simple function and } \phi \leq f \right\}

(We do allows $\int_E f , dm = \infty$)

For general measurable function f, we can define f^-(x)=\max\{0,-f(x)\}, f^+(x)=\max\{0,f(x)\}. (The positive part of the function and the negative part of the function, both non-negative)

Then f=f^+-f^-.

We say f is integrable if \int_E f^+ \, dm < \infty and \int_E f^- \, dm < \infty. (both finite) If at least one is finite, define


\int_E f \, dm = \int_E f^+ \, dm - \int_E f^- \, dm

We allow for A-\infty = -\infty and A+\infty = \infty for any A\in \mathbb{R}. But not \infty-\infty.

Immediate Properties of Lebesgue Integral

If f is measurable and m(E)=0, then \int_E f \, dm = 0.

If E=E_1\cup E_2 and E_1\cap E_2=\emptyset, then \int_E f \, dm = \int_{E_1} f \, dm + \int_{E_2} f \, dm.

Corollary

If f\leq g almost everywhere, (f\leq g except for a set of measure 0), then \int_E f \, dm \leq \int_E g \, dm.

Proof:

Let F=\{x\in E: f(x)>g(x)\}. Then m(F)=0.


\begin{aligned}
\int_E f \, dm &= \int_{E\setminus F} f \, dm + \int_F f \, dm\\
&\leq \int_{E\setminus F} g \, dm+\int_E g \, dm\\
&= \int_E g \, dm
\end{aligned}

QED

Proposition 6.13

If f is non-negative and \int_E f \, dm =0, then f=0 almost everywhere on E, f(x)=0 \forall x\in E\setminus F, where m(F)=0.

Proof:

Let E_n=\{x\in E: f(x)\geq \frac{1}{n}\}. Then \frac{1}{n}\chi_{E_n}(x)\leq f(x) for all x\in E.

By definition \frac{1}{n}m(E_n)=\int_E \frac{1}{n}\chi_{E_n} \, dm \leq \int_E f \, dm =0.

Therefore, m(E_n)=0 for all n.

Now U=\{x\in E: f(x)>0\}=\bigcup_{n=1}^{\infty} E_n.

QED

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