I am curious about the Dominated Convergence Theorem for a sequence of functions that converges in measure.


Let $(X,\mathcal{S},\mu)$ be a measure space. If $\{f_n\}, f$ are measurable, real-valued (i.e. finite) and such that $f_n \to f$ in measure and $|f_n| \leq g$ with $\int g \,\mathrm{d}\mu < \infty$, then $$ \int f_n \,\mathrm{d}\mu \to \int f \,\mathrm{d}\mu $$


Since $f_n \to f$ in measure, then for any subsequence of $\{f_n\}$, call it $\{f_k\}$, we have also $f_k \to f$ in measure. Now we can extract a further subsequence $\{f_{k_j}\}$ such that $f_{k_j} \to f$ almost everywhere. Applying the a.e. version of dominated convergence to this subsequence gives: $$ \int f_{k_j} \,\mathrm{d}\mu \to \int f \,\mathrm{d}\mu $$ Now defining the sequence of real numbers $\{a_n\}$ by $a_n = \int f_n \,\mathrm{d}\mu$, we want to show that $$a_n \to \int f \,\mathrm{d}\mu$$ But we have just shown that for any subsequence $\{a_k\}$ of $\{a_n\}$, there exists a further subsequence $\{a_{k_j}\}$ that converges to $\int f \,\mathrm{d}\mu$. Thus $a_n$ converges to $\int f \,\mathrm{d}\mu$ as well. QED.


So, nowhere in this proof did I use the fact that $\mu$ is $\sigma$-finite. However, everywhere that I look, I keep seeing this result with the condition that $\mu$ be $\sigma$-finite (e.g: Generalisation of Dominated Convergence Theorem). So I must be doing something wrong?

The only place I can think of where $\sigma$-finiteness might be required is in extracting an a.e. convergent subsequence from the "in measure" convergent sequence $\{f_k\}$. But I am pretty sure that $\sigma$-finiteness is not required to extract an almost uniformly convergent subsequence from an "in measure" convergent subsequence. And since almost uniformly convergent subsequences are also almost everywhere convergent, then I'm stumped.

Any pointers?

  • $\begingroup$ I edited the conditions on $g$: otherwise it could happen that it's a negative function with big absolute value whose integral would certainly be less than infinity. $\endgroup$
    – Ilya
    Jun 24, 2013 at 11:32
  • $\begingroup$ Besides that, the proof of DCT and the existence of a.e. convergence subsequence indeed seem not to require $\sigma$-finiteness. Why do you think it is required? $\endgroup$
    – Ilya
    Jun 24, 2013 at 11:39
  • 6
    $\begingroup$ Everything of interest happens in the $\sigma$-finite set $\{x : g(x) \gt 0\}$ anyway, so proving the $\sigma$-finite case is enough to establish the seemingly more general case. $\endgroup$
    – Martin
    Jun 24, 2013 at 14:15
  • $\begingroup$ @Ilya: Thanks for the edit. I just thought it was required because I kept seeing it stated with the requirement for $\sigma$ finiteness, in various places. But when I worked it out, it didn't seem to require it. So essentially, the a.e. statement of DCT holds exactly as is, even if "a.e." is replaced with "in measure?" I was just trying to confirm my suspicions... $\endgroup$
    – gogurt
    Jun 24, 2013 at 14:35
  • $\begingroup$ seems so, see also the enlightening comment by @Martin $\endgroup$
    – Ilya
    Jun 24, 2013 at 14:39

1 Answer 1


Some remarks:

  1. One can prove (classical) dominated convergence theorem from Egorov's theorem when the measure space is finite.

  2. Then even when the measure space is not a finite union of measurable sets of finite measure, we can work on $(X_N,X_N\cap\mathcal B,\mu\mid_{X_N\cap\mathcal B})$, where $X_N=\{g>N^{-1}\}$. (we use a $2\varepsilon$-argument, choosing first $N$ such that $\int (g-g\chi_{X_N})<\varepsilon$).


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