Is a $L^p$ function almost surely bounded a.e.?

Question

I just have a quick question related to $L^p$ spaces. Any help is greatly appreciated.

Is it true that if a function $f$ belongs to $L^p$ space, absolute value of $f$ raise to the power of $p$ is bounded a.e?

My guess is yes, I just wanted to confirm...

Thank you.

Answer 1

Yes if $p=\infty$, by definition. No for $1\le p<\infty$, in general. David Mitra already gave a counterexample.

Here is a more precise statement. Let $(X,\mu)$ be a measure space, and fix $p\in [1,\infty)$. The following are equivalent:

1. For every $f\in L^p(X,\mu)$ the essential supremum of $|f|$ is finite.
2. There exists $\delta>0$ such that for every measurable set $E\subseteq X$ either $\mu(E)=0$ or $\mu(E)\ge \delta$.

The standard example of a measure space that satisfies 2 is $\mathbb N$ with the counting measure. In this setting, $L^p$ becomes the sequence space $\ell^p$.

Proof. Suppose 2 fails. Then there is a sequence $E_n$ of sets such that $0<\mu(E_n)<2^{-n}$ for all $n$. Let $f=\sum n \chi_{E_n}$. Then $f$ has infinite essential supremum, and $$\|f\|_{L^p}\le \sum_n n\|\chi_{E_n}\|_{L^p} < \sum_n n\, 2^{-pn}<\infty$$

Conversely, suppose 2 holds. Given $f\in L^p$, choose $t>\delta^{-1/p}\|f\|_{L^p}$ and consider the set $E=\{x: |f(x)|>t\}$. If $\mu(E)>0$, then $$\|f\|_{L^p}^p \ge t^p\mu(E)>\delta^{-1} \|f\|_{L^p}^p\,\delta =\|f\|_{L^p}^p$$ which is a contradiction. Thus, $|f|\le t$ a.e.