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Can you tell me where the mistake is?

If $(f_n) \in L^1$ is a sequence of functions such that $\sum_n \|f_n\|_1 < \infty$ I can prove that $f(x) = \sum_{n=1}^\infty f_n(x) < \infty$ for all $x \in X$.

To this end, let $n \in \mathbb N$, $x \in X$. Then $$ \sum_{k=1}^n f_k (x) \leq \sum_{k=1}^n |f_k (x) | \leq \sum_{k=1}^n \int_X |f_k (x) | d \mu$$

Hence $$ f(x) = \lim_{n \to \infty} \sum_{k=1}^n f_k (x) \leq \lim_{n \to \infty} \sum_{k=1}^n \int_X |f_k (x) | d \mu < \infty$$

Since $x$ was arbitrary we have $f(x) < \infty$ for all $x \in X$.

This proof must be wrong since we should have $f(x) < \infty$ almost everywhere.

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Where does the inequality $\sum_{k=1}^n |f_k (x)| \leq \sum_{k=1}^n \int_X |f_k (x)| d \mu$ comes from? In general, it is false. – D. Thomine Jul 4 '12 at 9:58
up vote 2 down vote accepted

In general it is not true that $$ |f(x)|\leq\Vert f\Vert_1=\int\limits_{X}|f(x)|d\mu $$ For counterexample consider function $$ f(x)= \begin{cases} 1&\quad\text{ if }\quad x=0\\ 0&\quad\text{ if }\quad x\neq 0 \end{cases} $$ then you get $$ |f(0)|=1>0=\int\limits_{\mathbb{R}} |f(x)|d\mu $$

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Which measure are you taking? – Davide Giraudo Jul 4 '12 at 11:45
@DavideGiraudo, it is Lebesgue measure – Norbert Jul 4 '12 at 11:51
In this case, you are cheating, because $f$ is equal almost eveywhere to $0$. You can take $f(x)=2\chi_{(0,1/2)}$. – Davide Giraudo Jul 4 '12 at 12:01
Well if we are talking about equivalence classes of functions yes I'm cheating. If we are talking about concrete functions, I'm not. But think a little. If we are talking about equivalence classes, then the notion of the value of the function at particular point is meaningless. So I assume that we deal with ususal functions. – Norbert Jul 4 '12 at 12:54

The mistake is when you claim $\sum_1^n{ |f^k(x)| } \leq \sum_1^n{ \int_X{|f^k(x)|d\mu}} $.

Why should this be true ? But i believe you can prove that $\sum_1^n{ |f^k(x)| }$ converges for almost every $x$, at least if $X$ is regular enough. Ask me if you want more precision

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