# Monotone Convergence Theorem for non-negative decreasing sequence of measurable functions

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 $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$

Atempt:

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 $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$

• 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 $g_n = f_1 - f_n$? Mar 24, 2012 at 20:06
• @cardinal: oh yes....$g_n \geq 0$...Thanks
– Kuku
Mar 24, 2012 at 20:07
• Yes, $g_n \geq 0$...and, what else? Davide's answer lays out the details. (+1 for showing your work.) Mar 24, 2012 at 20:08
• Fair enough. Sorry, being a "standard" result, it sounded a bit like homework. Cheers. :) Mar 24, 2012 at 20:18
• Why not invoking the dominated convergence theorem with dominating function $f_1$? Aug 9, 2017 at 13:24

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 $$n$$.
• I used it with $f_1$ not $f$. Jan 17, 2014 at 10:17
• @Ale. $f_1\geq f_n$, since $f_n$ being decreasing and hence $f_1-f_n\geq 0$ for each $n$. Jun 30, 2016 at 5:44
• 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.
• @ZFR Since $f_1$ is integrable, $f_1<\infty$ almost everywhere hence $0\leq f_n<\infty$ almost everywhere. Aug 5, 2020 at 16:01