Can we say that : $ \exists f_{\infty}\in L_{\mathbb{R}}^{1} \text{ such that: } f_n\to f_\infty\text{ a.e and in } L_{\mathbb{R}}^{1} $ [closed]

Question

Let $(E,\mathcal{A},\mu)$ be a finite measure space and $\{f_n\}\subset L_{\mathbb{R}}^{1}$ such that: $$ \sum_{i=2}^{\infty}{\int_{E}{|f_n(t)-f_{n-1}(t)|d\mu(t)}}<+\infty $$ Can we say that : $$ \exists f_{\infty}\in L_{\mathbb{R}}^{1} \text{ such that: } f_n\to f_\infty\text{ a.e and in } L_{\mathbb{R}}^{1} $$

An idea please.

Answer 1

Assuming that $L^1_{\mathbb R}$ denotes the space of all classes of $\mu$-equivalent functions $f\colon E\to\mathbb R$ with $\|f\|_1=\int_E|f|\,d\mu<\infty$, the answer is yes.

Indeed, we have $$\sum_2^\infty\|f_i-f_{i-1}\|_1<\infty.$$ So, for natural $n>m$ $$\|f_n-f_m\|_1\le\sum_{m+1}^n\|f_i-f_{i-1}\|_1\to0$$ as $m\to\infty$. Therefore and because $L^1_{\mathbb R}$ is complete (see e.g. this), for some $f_\infty\in L^1_{\mathbb R}$ we have $$\|f_n-f_\infty\|_1\to0$$ as $n\to\infty$.

Moreover, for natural $k$ $$h_k:=\sup_{n\ge k}|f_n-f_\infty|\le\sum_{n\ge k}|f_n-f_{n-1}|,$$ whence $$\|h_k\|_1\le\sum_{n\ge k}\|f_n-f_{n-1}\|_1\to0$$ as $k\to\infty$. So, $h_k\to0$ in measure $\mu$ as $k\to\infty$ -- that is (cf. e.g. Two equivalent definitions of almost sure convergence), $f_n-f_\infty\to0$ as $n\to\infty$ $\mu$-almost everywhere.