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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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$\infty / \infty=?$ – Michael Greinecker Oct 29 '12 at 1:05
I know, that's my problem! But can't I cancel them with each other? – Anita Oct 29 '12 at 1:07
You can probably rewrite the problem so that it does not occur. But the expression is simply meaningless. – Michael Greinecker Oct 29 '12 at 1:11
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?) – Berci Oct 29 '12 at 1:39
What if $f_n \rightarrow f$ pointwise? – Anita Oct 29 '12 at 4:45

Here is an example of what has been explained to you in the comments: $X=(0,1)$, $\mathcal F=\mathcal B(X)$, $\mu=\mathrm{Leb}$, $f_n:x\mapsto1/(n\sqrt{x})$. You might want to see what happens to the statements you try to prove, in this case. First, what is the function $f$? Next, what is $\|f_n-f\|_\infty$? Hence...

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