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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