Show that $\int fd\mu=\infty$ for measurable sets $A$ and $0$ whenever $\mu(A) = 0$

Question

Let $f$ be a nonnegative function. Show that $\int fd\mu=\infty$ if there exists a measurable set $A$ with $\mu(A)>0$ such that $f(x)=\infty$ for $x∈A$. On the other hand, Show $\int fd\mu=0$ for $f$ of the form $f(x)=\infty \chi_A(x)=\begin{cases} \infty & x\in A\\ 0 & x\notin A\\ \end{cases}$ whenever $\mu(A) = 0$.

I started with $$\int fd\mu=\int\sum_{k=1}^n\alpha_k\chi_{A_n}d\mu$$ and got stuck. is it simply? $$\int\sum_{k=1}^n\alpha_k\chi_{A_n}d\mu=\begin{cases} \infty & x\in A\\ 0 & x\notin A\\ \end{cases}$$

I am at a total loss with this.

Answer 1

Let $X$ be the set on which you have your measure $\mu$. ($X$ is the entire domain $f$.) I am guessing that, by $\int f d \mu$ you mean $\int_{X} f d \mu$. So, since $A \subset X$, we have $$ \int f \; d\mu \geq \int_{A} f \; d\mu = \infty \, \mu(A) = \infty, $$ where the last equality holds because $\mu(A) > 0$.

If now $\mu(A) = 0$ and if $f = 0$ outside of $A$, then $$ \int f \; d\mu = \int_{A} f \; d\mu = \infty \, \mu(A) = \infty $$ by convention $0 \cdot \infty = 0$, and you get your other result.

Answer 2

How do we use the fact that $f(x)=\infty$ for all $x\in A$? One way: It means for any $M>0$, $f(x)>M$ for all $x\in A$. We can use this observation and monotonicity of the integral (i.e. If $f(x)\ge g(x)$ for all $x\in A$, then $\int_A f\ge \int_A g$) to bound $\int_A f$ below by a constant function that is high as we want. In particular, we know that $f(x)\ge g(x)= M/\mu(A)$ for all $x\in A$. So we get $\int_A f\ge \int_A M/\mu(A)=M$. Since $M$ is an arbitrary positive number, we've shown $\int_A f$ is arbitrarily large.

I'm not sure if this shows $\int f=\infty$; what if there's another set $B$ with positive measure so that $f(x)=-\infty$ for all $x\in B$? You may need another condition like $\int_{A^c} |f|<\infty$.