TITLE: How "bounded" are $L^1$ functions?
QUESTION [5 upvotes]: I am well aware of the fact that $L^1-$functions are not necessarily essentially bounded. Take for instance the function $1/\sqrt{x}$ on $X=(0,1)$.
However, can we say that they are "almost" bounded in the sense that if we cut out the bad parts with an epsilon of room they are bounded a.e.? Formally:

Suppose $(X,\mu)$ is a measure space and $f\in L^1(X)$. Then for every $\varepsilon>0,$ there exists a set $F\subset X$, such that $\mu(F)<\varepsilon$ and $f$ is essentially bounded on $X\setminus F$.

Does this hold? I can't think of any counterexample. The usual ones to show that $L^1$ is not necessarily bounded clearly don't work (for instance above we can surely cut out any $\epsilon$ segment around $0$).

REPLY [3 votes]: Actually, this holds for arbitrary (real or complex valued) measurable functions, not just for integrable ones.
Simply note that
$$
X = \bigcup_n |f|^{-1}([0,n]),
$$
where the sets in the union increase with $n$.
By continuity of the measure from below, we see that (since $\mu(X) < \infty$), there is some $n \in \Bbb{N}$ with $\mu(|f|^{-1}([0,n])) > \mu(X) - \varepsilon$, which shows that $\mu(|f|^{-1}((n,\infty)) < \varepsilon$.