Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X,\mathcal{M},\mu)$ be a measure space. 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 $$\int_X f~d\mu = \lim_{n\to\infty}\int_X f_n~d\mu.$$

share|improve this question
1  
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 $g_n = f_1 - f_n$? –  cardinal Mar 24 '12 at 20:06
    
@cardinal: oh yes....$g_n \geq 0$...Thanks –  Kuku Mar 24 '12 at 20:07
    
Yes, $g_n \geq 0$...and, what else? Davide's answer lays out the details. (+1 for showing your work.) –  cardinal Mar 24 '12 at 20:08
    
@Cardinal...Is not homework. I saw it being used here: math.stackexchange.com/questions/86676 and thought I might try and prove it. –  Kuku Mar 24 '12 at 20:16
1  
Fair enough. Sorry, being a "standard" result, it sounded a bit like homework. Cheers. :) –  cardinal Mar 24 '12 at 20:18

1 Answer 1

up vote 10 down vote accepted

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 $f_1-f_n$. 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 $\int_{\mathbb R}f_nd\lambda =+\infty$ for all $n$.

share|improve this answer
    
(+1) I was composing my comment simultaneously. We even managed to match notation. :) –  cardinal Mar 24 '12 at 20:07
    
Thanks again...I should have thought about it more:) –  Kuku Mar 24 '12 at 20:19
    
@cardinal Nice proof, yet I have a question: You say that $g_n:= f-f_n$ is an increasing sequence of function, I understand this, but why is she positive? I've thought that since $f_n$ is decreasing to f than $f_n \ge f \forall n$ but this would mean that $f-f_n$ is negative. Where is my error? I can't find it Thank you in advance if you answer :-) –  Ale Jan 17 at 7:53
    
I used it with $f_1$ not $f$. –  Davide Giraudo Jan 17 at 10:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.