# Thread:

1. ## Sequence, Lebesgue Space

Show that if $|| f_n ||_1 \leq 2^{-n}$ for every $n \geq 1$ then $(f_n)_n$ converges to zero a.e.

If $(f_n)_n$ is a sequence in $L^1[0, 1]$ is such that $\sum_{n=1}^{\infty} || f_n ||_1 < \infty$ then $\sum_{n=1}^{\infty} | f_n(s) | < \infty$ for almost every $s \in [0, 1]$.

However, I don't see how to show this. I would appreciate some hints on how to proceed.

2. Originally Posted by eskimo343
Show that if $|| f_n ||_1 \leq 2^{-n}$ for every $n \geq 1$ then $(f_n)_n$ converges to zero a.e.

If $(f_n)_n$ is a sequence in $L^1[0, 1]$ is such that $\sum_{n=1}^{\infty} || f_n ||_1 < \infty$ then $\sum_{n=1}^{\infty} | f_n(s) | < \infty$ for almost every $s \in [0, 1]$.

However, I don't see how to show this. I would appreciate some hints on how to proceed.
Not true, take $f_n(x)=1$ if $x=0$ and $f_n(x)=0$ otherwise. Then $\sum\parallel f_n\parallel_1<\infty$ but $\sum|f_n(0)|=\infty$

3. Originally Posted by putnam120
Not true, take $f_n(x)=1$ if $x=0$ and $f_n(x)=0$ otherwise. Then $\sum\parallel f_n\parallel_1<\infty$ but $\sum|f_n(0)|=\infty$
That does not contradict the assertion that $\sum_{n=1}^{\infty} | f_n(s) | < \infty$ for almost every $s \in [0, 1]$.

Originally Posted by eskimo343
Show that if $|| f_n ||_1 \leq 2^{-n}$ for every $n \geq 1$ then $(f_n)_n$ converges to zero a.e.
Let $g_N = \sum_{n=N+1}^\infty|f_n|$. Then $\|g_N\|_1\leqslant\sum_{n=N+1}^\infty\|f_n\|_1\leq slant2^{-N}$ (sum of geometric series). Define $S_N = \{x:g_N(x)\geqslant2^{-N/2}\}$. Then $\mu(S_N)\leqslant2^{-N/2}$, where $\mu$ denotes the measure.

If $x\notin S_N$ then $|f_n(x)|<2^{-N/2}$ for all n>N. Therefore if $x\notin\limsup_{N\to\infty}S_N$ then $f_n(x)\to0$ as $n\to\infty$. But $\limsup_{N\to\infty}S_N$ is a null set.
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