So I had this question on a recent test and I wasn't able to do it.

Let $g(x)$ be integrable and let $\{f_n\}$ be a sequence of measurable functions such that $|f_n|\leq g(x)$ and $f_n\rightarrow f$ almost everywhere. I want to show that $\int |f_n-f|\rightarrow 0$.

I think the solution some how involves the Lebesgue convergence theorem, but I don't know how to apply it. I could use some help.


  • $\begingroup$ Isn't this actually the statement of the Lebesgue Dominated Convergence Theorem? $\endgroup$
    – JT_NL
    Oct 30, 2011 at 18:50
  • $\begingroup$ Or at most a rephrasing. $\endgroup$
    – JT_NL
    Oct 30, 2011 at 19:06

1 Answer 1


You can argue like that

$|f_n-f|\to 0$ a.e. by hypothesis. Since $|f_n(x)|\leq g(x)$ for all $n$ and $f_n$ converges to $f$, a.e. we have that $|f(x)|\leq g(x)$ a.e. By the triangular inequality we get the bound $|(f_n-f)(x)|\leq 2 g(x)$. Now we can use the Lebesgue Dominated Convergence Theorem.

  • $\begingroup$ Huh! really straight forward, once you establish the bounds. Thanks $\endgroup$
    – Kuku
    Oct 30, 2011 at 19:17
  • $\begingroup$ You are welcome Kuku. $\endgroup$
    – Leandro
    Oct 30, 2011 at 19:23
  • $\begingroup$ @Leandro Sorry to bother, but how can we explicitly show that $|f(x)|\leq g(x)$ from those two assertions? It seems obvious, but I'm hoping for a formal reason? thanks :) $\endgroup$ Oct 30, 2011 at 20:11
  • $\begingroup$ @aengle: consider each $x$ for which $\lim\limits_n f_n(x) = f(x)$. Then this assertion is obvious since it is a numerical limit $\endgroup$
    – SBF
    Oct 30, 2011 at 20:23

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