TITLE: Locally integrable function with a uniform bound...
QUESTION [5 upvotes]: I'm a bit lost...  I have a measure space $(\Omega,\mathcal{B}(\Omega),\mu)$  where $\mathcal{B}(\Omega)$ is a Borel set. Let $f$ be a real-valued measurable function on $\Omega$ and $\mathcal{K}$ be the set of all compact sets in $\Omega$. If
$$
\sup_{K\in\mathcal{K}}\int_K \lvert f(x) \rvert\, \mu(dx) < +\infty\,, 
$$
do we have $f\in L^1(\Omega,\mu)$?
We can add any nice property on $\Omega$ like locally compact, or $\mu$ being $\sigma$-finite, etc. Thanks in advance... Maybe a hint because it is not clear at  all for me. 
Sincerely.

REPLY [5 votes]: It suffices to know that there is a sequence of compact sets $K_n$ that satisfies $\mu(\Omega \setminus \bigcup K_n) = 0$ (why!? Use monotone convergence and the fact that the $K_n$ can be taken to be an increasing sequence). 
The above is fulfilled e.g. if
1) $\Omega$ is $\sigma$-compact, or
2) $\mu$ is $\sigma$-finite and inner regular, because then $\Omega = \bigcup M_n$ with an increasing sequence $M_n$ each of finite measure, so that for every $n$ there exists a compact $K_n$ with $\mu(M_n \setminus K_n) < 2^{-n}$, so that
$$
\mu (\Omega \setminus \bigcup K_n) = \lim_m \mu(M_m \setminus K_n) \leq \lim_m 2^{-m} = 0. 
$$