For measurable $f_n : X \to [0, \infty)$ show that $\sum_{n=1}^\infty f_n < \infty$ almost everywhere

Question

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 $x \in X$.

Answer 1

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 $f_n(x)\leqslant M$.

Answer 2

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).