Properties of Lebesgue Integration

Question

I am completely stuck with the following problem on Lebesgue Integration:

Let $f:\mathbb{R}^d \to [0, +\infty]$ be measurable.

1. Show that if $\int_{\mathbb{R}^d} f(x)dx < \infty$, then $f$ is finite almost everywhere. Give a counterexample to show that the converse statement is false.
2. Show that $\int_{\mathbb{R}^d} f(x)dx=0$ if and only if $f$ is zero almost everywhere.

All I know is the definition of a measurable function:

An unsigned function $f:\mathbb{R}^d \to [0, +\infty]$ is unsigned Lebesgue measurable, or measurable for short, if it is the point-wise limit of unsigned simple functions, i.e., if there exists a sequence $f_1,f_2,f_3, \ldots : \mathbb{R}^d \to [0, +\infty]$ of unsigned simple functions such that $f_n(x) \to f(x)$ for every $x \in \mathbb{R}^d$.

How would I be able to use this to my advantage in order to solve the problem at hand? Thank you in advance for helping me with these questions.

Answer 1

To prove your statements we will use the following

Proposition (Markov inequality): Let $f$ be a positive measurable function, then for all $a > 0$ we have $$\mu\{x \in E: f(x)>a \}\le\frac{1}{a}\int fd\mu$$ Where $E$ is a measurable set where $f$ is defined.

Proof

Define $A_a:=\{ x\in E:f(x)>a\}$ (that is the set we want to measure). Then we see that

$$f(x) \ge a\Bbb{1}_{A_a}$$

Where $\Bbb{1}_{A_a}:= \begin{cases} 1, & \text{ if } x \in A_a \\0, &\text{else}\end{cases}$. Indeed if $x \in A_a$ then we have $f(x) \ge a$ by definition of $A_a$ and otherwise $f(x) \ge 0$, which is true since $f$ is a positive function. Now integrate both sides to get

\begin{align*}\int f d\mu &\ge a\int \Bbb{1}_{A_a} d\mu \\ &\ge a\mu (A_a) \implies \mu(A_a) \le\frac{1}{a} \int f d\mu\end{align*} as we wanted.

Now to prove the statement

$$\int_E fd\mu \lt \infty \implies f \lt \infty$$ if $f$ is positive and measurable

what we want to do is to measure the set where $f$ is not finite. Ideally we want that set to have measure $0$, so that we know that our function does not take infinite values.

Define then for all $n \ge 1$, $A_n:= \{x \in E:f(x) \ge n \}$, $A_{\infty}:=\{x \in E: f(x) = \infty \}$. Note that $A_{\infty}$ is the intersection over all $n$ of $A_n$. Hence we have

$$\mu \left(A_{\infty}\right)= \mu \left(\bigcap_{n\ge 1}A_n\right)=\lim_{n\to \infty} \mu(A_n) \le \lim_{n \to \infty} \frac{1}{n} \int f d\mu\to 0\qquad(n\to \infty)$$

Where we first use the fact that the intersection of the $A_n$ is a decreasing intersection and where we used the proposition above to approximate the measure of $A_n$. Now you are done, since the measure of $A_{\infty}$ is zero and therefore you know that $f$ does not take infinite values.

To prove the statement

for $f$ positive and measurable $$\int f d \mu = 0 \iff f =0\qquad a.e$$

You need to prove two directions. If you have $f=0$ almost everywhere then it's quite easy, but if you don't know how to do it ask. The other direction is a little bit more difficult, but as above the idea is to measure the set where $f$ takes value bigger than 0. We want that set to have measure $0$. Define then for all $n \ge 1$ $B_n := \{ x \in E: f(x) \ge \frac{1}{n}\}$. Again using the proposition above we get

$$\mu(B_n) \le n \underbrace{\int f d\mu}_{=0}$$