# A measurable function with $\int f^n$ bounded or converging as $n \to \infty$

(1) Show that, if $f^n$ is integrable for all integers $n\ge 1$ and $\limsup_{n\to \infty} \int f^n<\infty$ then $|f|\le1$ almost everywhere.

(2) Show that, if $f^n$ is integrable for all integers $n\ge 1$ then $\int f^n d\mu = c$, for every $n\in\mathbb N$ $\Leftrightarrow f=\mathbb{1}_{A}$ a.e. for some $A\subseteq X$ with $\mu(A)=c$. (This is the edited version, i think here is not necessarily $|f|\le 1\ a.e$)

Proof(1):

Assume not ( $|f|\le 1$ not almost everywhere)

Since $||f||_{\infty}=\inf\{M:|f(x)|\le M\quad \text{for } \mu \text{-almost everywhere } x\in X\}$

$\Rightarrow ||f||_{\infty}>1$ then $||f||_{\infty}^{\infty}=\infty$ which is a contradiction.

Is this OK?

• Your attempt would be correct if you make a link between the supremum of integrals of $n$-th powers of $f$ and the supremum norm. – Davide Giraudo Nov 14 '13 at 13:13

## 1 Answer

(1) Assume that $|f|\leqslant 1$ doesn't hold almost everywhere. Then there is $k\geqslant 1$ such that $\mu(A)\gt 0$, with $A:=\{|f|\geqslant 1+k^{-1}\}$. We thus have $$\int_X|f|^n\mathrm d\mu\geqslant \int_A|f|^n\mathrm d\mu\geqslant \mu(A)(1+k^{-1})^n,$$ which contradicts the finiteness of $\sup_n\int_X|f|^n\mathrm d\mu$.

(2) One direction is easy. For the other one, we have to prove that $f\in \{0,1\}$ almost everywhere. Thanks to the first part, it's enough to show that $\mu\{0\lt f\lt 1\}=0$. To this aim, define $A_k:=\{k^{-1}\leqslant f\leqslant 1-k^{-1}\}$.

• ok then, $\mu(\{0<|f|<1\})=0\Leftrightarrow \int\limits_{(\{0<|f|<1\})}f^n=\int\limits_{\bigcup A_k} f^n\le\sum\int\mathbb 1_{A_k}f^n\le\sum\int\underbrace{\mathbb 1_{A_k}(1-k^{-1})^n}_{0,\ if n\rightarrow\infty\forall k}=0$ correct ? – derivative Nov 20 '13 at 12:44