Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f_n : X \to [0 \infty)$ be a sequence of measurable functions on the measure space $(X, \mathcal{F}, \mu)$. Suppose there is an $M > 0$ such that the functions $g_n = f_n\chi_{\{f_n \le M\}}$ satisfy $||g_n||_1 \le An^{-\frac{4}{3}}$ and for which $\mu\{f_n > M\} \le Bn^{-\frac{5}{3}}$. Here, $A$ and $B$ are positive constants independent of $n$. Prove that $h(x) = \displaystyle \sum_{n=1}^\infty f_n(x) < \infty$ for almost all $x \in X$.

share|cite|improve this question

As the sequence $\{\sum_{n=1}^Nf_n\chi_{\{f_n\leqslant M\}}\}$ is convergent in $L^1$, we extract an almost everywhere convergent sequence. As the concerned terms are non-negative, we actually have that $\sum_{n=1}^{+\infty}f_n\chi_{\{f_n\leqslant M\}}$ is convergent for almost everywhere $x$.

By a Borel-Cantelli like argument, $\mu(\limsup_{n\to+\infty}\{f_n>M\})=0$, so for almost every $x$, we can find an integer $N(x)$ such that if $n\geqslant N(x)$ then $f_n(x)\leqslant M$.

share|cite|improve this answer

The Borel-Cantelli Lemma and the bound on each $\mu\{f_n > M\}$ ensure that $\mu\{x \in X : f_n(x) > M \mbox{ infinitely often}\} = 0.$ Thus, for almost all $x \in X$, there is an $N(x) \in \mathbb{N}$ so that $ \displaystyle \sum_{n=N(x)}^\infty f_n(x) =\sum_{n=N(x)}^\infty f_n(x)\chi_{\{f_n > M\}}(x)$. But $\displaystyle \sum_{n=1}^\infty f_n(x)\chi_{\{f_n > M\}}(x) < \infty$ for almost all $x \in X$, since $\displaystyle \int_X\sum_{n=1}^\infty f_n\chi_{\{f_n > M\}}d\mu = \sum_{n=1}^\infty\int_X f_n\chi_{\{f_n > M\}}d\mu \le \sum_{n=1}^\infty An^{-\frac{4}{3}} < \infty$ (the equality here is a consequence of the Monotone Convergence Theorem).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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