(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 $|f|\le 1\ a.e$)


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?

  • 2
    $\begingroup$ 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. $\endgroup$ – Davide Giraudo Nov 14 '13 at 13:13

(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 $A_k:=\{k^{-1}\leqslant f\leqslant 1-k^{-1}\}$.

  • 1
    $\begingroup$ 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 ? $\endgroup$ – derivative Nov 20 '13 at 12:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.