Billingsley's Probability and Measure, Problem 15.1, Part 2 Background:
Consider a measure space $(\Omega,{\mathscr F},\mu)$ and a $\mu$-measurable function $f:\Omega\to[0,\infty]$. The upper integral is defined as
$$\int^*fd\mu=\inf\sum_i\left(\sup_{\omega\in A_i}f(\omega)\right)\mu(A_i),$$
where the infimum is taken over all finite partitions $\{A_i\}$ of $\Omega$ into ${\scr F}$-sets.
Problem (15.1, Part 2):
Show that $\int^*fd\mu=\infty$ if $\mu\{\omega:f(\omega)\geq x\}>0$ for all (finite) $x$.
What I've tried so far is to find a lower bound, in terms of $x$, for $\sum_i\left(\sup_{\omega\in A_i}f(\omega)\right)\mu(A_i)$ and show that this lower bound is not bounded above (as a function of $x$):
Fix an arbitrary finite $x$. Let $B_x=\{\omega:f(\omega)\geq x\}$. Note that for any finite partition $\{A_i\}$ of $\Omega$ into ${\scr F}$-sets, $\sum_i\left(\sup_{\omega\in A_i}f(\omega)\right)\mu(A_i)\geq \sum_ix\mu(A_i\cap B_x)=x\mu(B_x)$.
However, this bound isn't strong enough; for instance if
$$\mu(B_x)=\begin{cases}e^{-x}&\text{(if $x\geq0$)}\\1&\text{(if $x<0$)}\end{cases}$$
then all we know is that $\int^*fd\mu\geq1/e$.
 A: You're not taking full advantage of the supremum. You can find a sequence of $x_i$ such that $f(x_i)$ diverges to infinity. Fix a finite partition $\{A_i\}$. Since the number of $A_i$ is finite, at least one must contain an infinite number of $x_i$, so that the supremum on that $A_i$ blows up.
A: We will show that if for all $x$, $\mu\{\omega:f(\omega)\geq x\}>0$, then for all finite partitions $\{A_i\}$ of $\Omega$ into ${\scr F}$-sets, there is an $i$ such that $f$ is unbounded on $A_i$ and $\mu(A_i)$>0. The result immediately follows from this fact, since we will then have $$\inf\sum_i\left(\sup_{\omega\in A_i}f(\omega)\right)\mu(A_i)\geq\left(\sup_{\omega\in A_i}f(\omega)\right)\mu(A_i)=\infty\cdot\mu(A_i)=\infty.$$
Consider any finite partition $\{A_i\}$ of $\Omega$ into ${\scr F}$-sets. It is sufficient to show the result for an arbitrary partition of an arbitrary subset of $\Omega$ into two ${\scr F}$-sets, since we can consider the partitions
$$A_1|A_1^c,~~~A_1|A_2|A_1^c\cap A_2^c,~~~\cdots,~~~A_1|\cdots|A_m,$$
result applying the result to
$$\Omega,~A_1^c,~A_1^c\cap A_2^c,~\cdots\text{ and }A_1^c\cap\cdots\cap A_{m-1}^c.$$
Say the sets are $A$ and $A^c$. First consider the case in which $\mu(A)\neq0$ and $\mu(A^c)\neq0$. Note that $f$ must be unbounded on $A$ or $A^c$, since it is unbounded on $\Omega$. On the other hand, if one of them has measure zero, then the result is obvious. Without loss of generality, say $A$ has measure zero. Then for all $x$,
$$\mu\{\omega\in A^c:f(\omega)\geq x\}=\mu\{\omega\in\Omega:f(\omega)\geq x\}>0,$$
so $f$ is unbounded on $A^c$. Note that $\mu(A^c)>0$ since $A^c\supseteq\{\omega\in A^c:f(\omega)\geq1\}$.
