Let $(X,\mathcal{M},\mu)$ be a measure space and suppose $\{f_n\}$ are non-negative measurable functions decreasing pointwise to $f$. Suppose also that $\int f_1 \lt \infty$. Then $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$


Since $\{f_n\}$ are decreasing, and converges pointwise to $f$, then $\{-f_n\}$ is increasing pointwise to $f$. So by the monotone convergence theorem $$ \int_X -f~d\mu = \lim_{n\to\infty}\int_X -f_n ~d\mu$$ and so $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$

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    $\begingroup$ Your attempt is on the right track but is not quite right. In particular, you might think about the hypothesis $\int f_1 < \infty$ and whether you've used it. Hint: What do you know about $g_n = f_1 - f_n$? $\endgroup$
    – cardinal
    Commented Mar 24, 2012 at 20:06
  • $\begingroup$ @cardinal: oh yes....$g_n \geq 0$...Thanks $\endgroup$
    – Kuku
    Commented Mar 24, 2012 at 20:07
  • $\begingroup$ Yes, $g_n \geq 0$...and, what else? Davide's answer lays out the details. (+1 for showing your work.) $\endgroup$
    – cardinal
    Commented Mar 24, 2012 at 20:08
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    $\begingroup$ Fair enough. Sorry, being a "standard" result, it sounded a bit like homework. Cheers. :) $\endgroup$
    – cardinal
    Commented Mar 24, 2012 at 20:18
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    $\begingroup$ Why not invoking the dominated convergence theorem with dominating function $f_1$? $\endgroup$
    – Mr_3_7
    Commented Aug 9, 2017 at 13:24

1 Answer 1


The problem is that $-f_n$ increases to $-f$ which is not non-negative, so we can't apply directly to $-f_n$ the monotone convergence theorem. But if we take $g_n:=f_1-f_n$, then $\{g_n\}$ is an increasing sequence of non-negative measurable functions, which converges pointwise to $f_1-f$. Monotone convergence theorem yields: $$\lim_{n\to +\infty}\int_X (f_1-f_n)d\mu=\int_X\lim_{n\to +\infty} (f_1-f_n)d\mu=\int_X f_1d\mu-\int_X fd\mu$$ so $\lim_{n\to +\infty}\int_X f_nd\mu=\int_X fd\mu$.

Note that the fact that there is an integrable function in the sequence is primordial, indeed, if you take $X$ the real line, $\mathcal M$ its Borel $\sigma$-algebra and $\mu$ the Lebesgue measure, and $f_n(x)=\begin{cases} 1&\mbox{ if }x\geq n\\ 0&\mbox{ otherwise} \end{cases}$ the sequence $f_n $ decreases to $0$ but $\int_{\mathbb R}f_nd\mu=+\infty$ for all $n$.

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    $\begingroup$ I used it with $f_1$ not $f$. $\endgroup$ Commented Jan 17, 2014 at 10:17
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    $\begingroup$ @Ale. $f_1\geq f_n$, since $f_n$ being decreasing and hence $f_1-f_n\geq 0$ for each $n$. $\endgroup$
    – Alexander
    Commented Jun 30, 2016 at 5:44
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    $\begingroup$ @AlwaysNeedHelp The last three lines of my post give an example of what could happen. $\endgroup$ Commented Nov 14, 2017 at 15:45
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    $\begingroup$ Dear Davide! I read your proof and it is quite nice but let me ask you a couple of question: 1) You defined $g_n:=f_1-f_n$ but here you can run into $\infty-\infty$, right? So you have to clarify this moment. 2) How do you know that $f_1-f_n$ converges pointwise to $f_1-f$? The issue that you can have $\infty-\infty$ here. Would be thankful if you can explain them. $\endgroup$
    – RFZ
    Commented Aug 5, 2020 at 15:35
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    $\begingroup$ @ZFR Since $f_1$ is integrable, $f_1<\infty$ almost everywhere hence $0\leq f_n<\infty$ almost everywhere. $\endgroup$ Commented Aug 5, 2020 at 16:01

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