How "bounded" are $L^1$ functions?

Question

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 $0$).

Answer 1

For $M>0$, set $E_M=\{x\in X:\lvert\,f(x)\rvert>M\}$. Then $$ \|\,f\|_{L^1(X)}=\int_X\lvert\,f\rvert\,d\mu\ge\int_{E_M}\lvert\,f\rvert\,d\mu\ge M\mu(E_M). $$ Therefore $$ \mu(E_M)\le\frac{\|f\|_{L^1(X)}}{M}, $$ while $f$ is absolutely bounded in the set $X\setminus E_M$ by $M$.

So, set $\displaystyle M=\frac{\|f\|_{L^1(X)}}{\varepsilon}$, and you are done!

Answer 2

Yes, definitely.

Let $A$ be any positive number, let $I(|f|)=C$, then $\mu\left( |f|^{-1}([A,\infty) )\right)<\frac{C}{A}$.

Make $A$ big, the measure must become tiny.

Answer 3

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 $\mu(|f|^{-1}((n,\infty)) < \varepsilon$.