Suppose that $f_{n}$ is a sequence of measurable functions, in a finite measure space, $f_{n}\to f $ in $m$-measure and that there exists $g$ in $L^1$ such that $\vert f_n\vert \le g$. Prove that $$ \lim_{n\to +\infty}\Vert f_n-f\Vert_{L^1}=0. $$

What I obviously thought of doing was splitting the difference $|f_n-f|$ to the less than and greater than $\epsilon$ and bound the greater part by $2g$. I am stuck right there, I can show it is finite but can not show it is less than epsilon.

Next I thought of using the R. Fisher's argument of getting the subsequence of $f_n$ which converges a.e, and finiteness of space give you a. uniform by Egoroff). But that way I can only show result will be good for the case of subsequence. I am not sure if I can conclude from there though( by arguing that original sequence and its subsequence goes to the same limit). I am sure I am missing something here. I would love to get out of this confusion. Help please.

  • 3
    $\begingroup$ The subsequence argument can be put to work. You could try proving that any subsequence of $(f_n)$ has a subsequence which converges to $f$ in $L^1$, then apply this lemma. $\endgroup$ Dec 25, 2012 at 0:07

1 Answer 1


Let $\epsilon > 0$ be given. Show there is $\delta > 0$ such that $\int_A (|g|+|f|) \ dm < \epsilon/3$ for any measurable set $A$ with $m(A) < \delta$.
Also there is a set $C$ such that $m(C) < \infty$ and $\int_{C^c} (|g| + |f|)\ dm < \epsilon/3$. Now since $f_n \to f$ in measure, for sufficiently large $n$ we have $|f_n - f| < \epsilon/(3 m(C))$ except on a set $B_n$ of measure $< \delta$. So for sufficiently large $n$, $$ \int |f_n - f| \ dm \le \int_{B_n} |f_n - f| \ dm + \int_{C^c} |f_n - f|\ dm + \int_C \frac{\epsilon}{3 m(C)} \ dm \le \dfrac{\epsilon}{3}+\dfrac{\epsilon}{3}+\dfrac{\epsilon}{3} =\epsilon $$

  • $\begingroup$ "Also there is a set $C$ such that $m(C) < \infty$ and $\int_{C^c} (|g| + |f|)\ dm < \epsilon/3$" Why? $\endgroup$
    – Deepak
    Dec 24, 2012 at 21:47
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    $\begingroup$ Actually you don't need this part since you're given that it is a finite measure space. But in any $\sigma$-finite measure space you can get it by the Dominated Convergence Theorem. $\endgroup$ Dec 24, 2012 at 23:12

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