# Thread: Need to prove |f(x)|<= 1 almost every where.

1. ## Need to prove |f(x)|<= 1 almost every where.

Let $f :X \rightarrow R$ be a measurable function in $X$ with $m(X)< \infty$.
Show that if $f^n$ is Lebesgue integrable for every $n$ and that $\lim_{n\to\infty}\int f^n\,dm$ exists in $R$,then $-1\le f(x)\le 1$ almost every where.

Can anyone help me with this problem??

2. Here is just a general idea. Assume that $|f|>1$ on some measurable set with $m(U)>0$. Then $f^n>M$ for any $M\in\mathbb{R}$ once $n$ is large enough. So what will this do to $\int f^ndm$?

Hint: $\int f^ndm=\int_U f^ndm+\int_{U^c}f^ndm$.

3. Just to elaborate a bit on putnam120's suggestion, for k=1,2,3,..., let $S_k = \{x\in X: |f(x)|\geqslant 1+1/k\}$. Then $\int_X\!|f^n|\,dm\geqslant \int_{S_k}\!\!|f^n|\,dm\geqslant (1+1/k)^nm(S_k)$. If $m(S_k)\ne0$ then this goes to infinity as $n\to\infty$. Also, if n is even then $|f^n| = f^n$, so $\lim_{n\to\infty}\int_X\!f^n\,dm$ will not exist in that case.

Therefore $m(S_k) = 0$ for all k and hence $\textstyle m\left(\bigcup_{k}S_k\right) = 0$. But $\textstyle \bigcup_{k}S_k = \{x\in X:|f(x)>1\}$. Therefore $-1\leqslant f(x)\leqslant 1$ almost everywhere.