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Is what I'm doing valid if we don't have any information on boundedness of $f$ or $f_n$?

let $X$ be a finite measure space and $\{f_n\}$ be a sequence of nonnegative integrable functions, $f_n \rightarrow f$ a.e. on $ X$. This is true that $$\left|\lim_{n \rightarrow \infty}\int_X |f_n-f|d\mu\right| \leq \sup |f_n-f|\ \mu(X).$$

Set $X$ is under my control, so I want to say I can make it as small as a measure zero set, $\epsilon=\frac{\epsilon_0}{\sup\ |f_n-f|}$ in this case, and on that I don't need to care if $f_n \nrightarrow f$ because:

$$\left|\lim_{n \rightarrow \infty}\int_X |f_n-f|d\mu\right| \leq \sup |f_n-f| \times \frac{\epsilon_0}{\sup\ |f_n-f|}= \epsilon_0$$

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    $\infty / \infty=?$2012-10-29
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    I know, that's my problem! But can't I cancel them with each other?2012-10-29
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    You can probably rewrite the problem so that it does not occur. But the expression is simply meaningless.2012-10-29
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    If they are integrable, maybe the $L_1$ norm is better choice than $\sup$. (I guess in the 'almost everwhere' context, $\sup$ is meant the $L_\infty$ norm, no?)2012-10-29
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    What if $f_n \rightarrow f$ pointwise?2012-10-29

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