using sup of an unbounded function

Question

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:

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

Answer 1

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...