f integrable implies set where f is infinite is measure zero

Let $(X,\mathcal{M},\mu)$ be a measure space. Let $f:X\to [0,\infty]$ be a measurable function. Suppose that $\int_X f d\mu<\infty$ .

(a) Show that the set $\{x\in X:f(x)=\infty\}\subseteq X$ is of $\mu$ -measure 0. (Intuitively, this is quite obvious, but we need to prove it rigorously.)

(b) Show that the set $\{x\in X:f(x)\neq 0\}\subseteq X$ is $\sigma$ -finite with respect to $\mu$ . i.e. it is a countable union of measurable sets of finite $\mu$ -measure.

We may use Markov’s inequality, which turns out to be very useful in this question.

Proof: (a) Let $E_k=\{x\in X:f(x)\geq k\}$ , where $k\in\mathbb{N}$ . Denote $E_\infty=\{x\in X:f(x)=\infty\}$ .

$E_K \downarrow E_\infty$ , and $\mu (E_1)\leq\frac{1}{1}\int_X f d\mu<\infty$ . (Markov Inequality!)

Then

$\begin{aligned}\mu(E_\infty)&=\lim_{k\to\infty}\mu (E_k)\\ &\leq\lim_{k\to\infty}\frac{1}{k}\int f d\mu\ \ \ \text{(Markov Inequality)}\\ &=0 \end{aligned}$

Therefore, $\mu(E_\infty)=0$ .

(b) Let $S_k=\{x\in X:f(x)\geq\frac{1}{k}\}$ , $k\in\mathbb{N}$ .

$\{x\in X:f(x)\neq 0\}=\cup_{k=1}^\infty S_k$

Therefore, $\mu(S_k)\leq k\int f d\mu<\infty$ , and we have expressed the set as a countable union of measurable sets of finite measure.

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