# Assume $\sup_n \int_\Omega f_n \, d\mu < + \infty$. Does it follow $\sup_n f_n(x) < +\infty$ a.e.?

Let $f_n \colon \Omega \to \mathbb R$ be a sequence of measurable functions w.r.t to $\mu$, being $\mu$ a probability measure on $\Omega$. Assume $$\sup_n \int_\Omega f_n \, d\mu = M < + \infty.$$ Does it follow $$\sup_n f_n(x) < +\infty \qquad \mu\text{-a.e. } x \in \Omega?$$

I am puzzled and I am getting crazy about this. This is not homework, this is a sort of self-posed problem I end up while I was solving completely different stuff.

Could you please help me? I have been trying Lebesgue's dominated convergence theorem and Monotone convergence theorem but I failed. On the other hand, I tried also to construct a counterexample but I do not manage to figure out the "shape" of $f_n$... Thanks.

The answer is no. Try to construct counterexample by $\Omega = [0,1]$ with Lebesgue measure, using sequences of nonnegative functions $f_n$ supported on
$$[0,1], \left[0, \frac 12\right], \left[\frac 12, 1\right], \left[0, \frac 13\right], \left[\frac 13, \frac 23\right], \left[\frac 23, 1\right], \left[0, \frac 14\right]\cdots$$
so that $\int_0^1 f_n = 1$ for all $n$.
• Thanks, you are absolutely right. It was easy... The piecewise constant functions we construct have the property that $\sup_n f_n(x)=+\infty$ for every $x\in [0,1]$. Thanks a lot and sorry for the easy question. – Romeo Oct 9 '15 at 9:17