Let $f$ be a Lebesgue measurable function from $[0,1]\to\mathbb{R}$. Let $\mu$ be Lebesgue measure. Does there exist a measurable set $B$ with $\mu(B)>0$ and an $M>0$ such that for all $x\in B$, $f(x)<M$.
Obviously for a general measure this is not true, say $f=1/x$ with a point mass at $0$, but can be easily shown for an integrable function. Does anyone know if it is true more generally for measurable functions in the Lesbesgue case?
Let $B_n = \{x \in [0,1] \mid |f(x)| \leq n\}$.
Then $[0,1] = \bigcup_n B_n$ (why?) with $B_n \subset B_{n+1}$.
Now use "continuity of the measure from below" to conclude $\mu(B_n) \to \mu([0,1]) = 1$ to conclude that for every $1>\varepsilon > 0$, there is some $B_n$ with $\mu(B_n) > 1-\varepsilon > 0$. Hence, you can even come arbitraryly close to the measure of the unit interval.
The example with the point measure at zero is not really valid, because you did not define $f(0)$ (you can not define it as $\infty$, because you require $f : [0,1] \to \Bbb{R}$, i.e. only real values are allowed).
If you allow $\infty$ as a value, the proof above does not work anymore (why?).
Even the claim is false in that case; simply consider $f \equiv \infty$.
Take $$B_M = \left\{x\in[0,1] \colon |f(x)|<M\right\}.$$ If $f$ is not infinity almost everywher, then $\mu(B_M)>0$ for some $M$.
