# Lebesgue integral

• Dec 21st 2009, 06:53 PM
GTO
Lebesgue integral
let $f$ be a Lebesgue integrable function on [0,1], and let $\{g_n\}$ be a sequence of bounded measurable functions on [0,1] that converges uniformly to a function $g$. Prove that
$\int_{[0,1]}g_nfdx \rightarrow \int_{[0,1]}gfdx$.

help will be appreciated so much
• Dec 21st 2009, 11:04 PM
GTO
dominated convergence theorem
i dont see why $g_n$ has to converge uniformly to $g$. to me if $g_n \rightarrow g$, then $g_nf \rightarrow gf$(i am not sure how to show). and $|g_nf| \leq Mf$ since there exists $M$ such that $|g_n| \leq M$ and since $f$ is integrable, so is $Mf$.
by dominated convergence theorem, i think we get what we want.

can someone help me with this one?
• Dec 22nd 2009, 01:17 AM
Swlabr
Quote:

Originally Posted by GTO
let $f$ be a Lebesgue integrable function on [0,1], and let $\{g_n\}$ be a sequence of bounded measurable functions on [0,1] that converges uniformly to a function $g$. Prove that
$\int_{[0,1]}g_nfdx \rightarrow \int_{[0,1]}gfdx$.

help will be appreciated so much

Firstly, you will want to use the monotone convergence theorem. This gives us equality (your working with the Lebesgue integral and your lecturer is trying to show you how inadequate the Riemann integral is; the monotone convergence theorem doesn't work for Riemann, but it does for Lebesgue so they will tend to use it lots). You can use this here as you are in a compact interval and so a uniformly convergent sequence is non-decreasing.

So, can you prove that your sequence $g_nf$ uniformly converges to $gf$ if the sequence $g_n$ uniformly converges to $g$?
• Dec 22nd 2009, 01:37 PM
GTO
monotone convergence thm
doesnt a sequence has to be nonnegative to use monotone convergence thm?
• Dec 22nd 2009, 04:49 PM
putnam120
You can also use the dominated convergence theorem once you show that $g$ is bounded (which it is). I assume that when you say $f$ is integrable you mean that $\int_0^1|f|d\mu<\infty$.

Also at Swlabr how do you know that the sequence non-decreasing?
• Dec 23rd 2009, 01:31 AM
Swlabr
Quote:

Originally Posted by putnam120
You can also use the dominated convergence theorem once you show that $g$ is bounded (which it is). I assume that when you say $f$ is integrable you mean that $\int_0^1|f|d\mu<\infty$.

Also at Swlabr how do you know that the sequence non-decreasing?

I'm now a tad confused. I was sure I found a result which said that any uniformly convergent sequence in a compact space is non-decreasing. However, I cannot seem to find this now and in retrospect it doesn't seem to make that much sense - something can converge uniformly from above...
• Jan 8th 2010, 11:04 PM
GTO
Quote:

Originally Posted by putnam120
You can also use the dominated convergence theorem once you show that $g$ is bounded (which it is). I assume that when you say $f$ is integrable you mean that $\int_0^1|f|d\mu<\infty$.

Also at Swlabr how do you know that the sequence non-decreasing?

since $g_n \rightarrow g$ uniformly, for $\epsilon >0,$ there is $N$ such that for all $n \geq N$ and for all $x \in [0,1]$, we have $|g-g_n|< \epsilon$.
So $|g|<|g_n|+\epsilon \leq M_n + \epsilon$ for all $n \geq N$. so $g$ is bounded. but how can i used the dominated convergence theorem?
• Jan 9th 2010, 05:01 PM
putnam120
Let $\epsilon>0$, since $g_n\to g$ uniformly there exist N such that $|g_n-g_N|<\epsilon$ for $n>N$. Thus $|g_n|\le\epsilon+|g_N|\le\epsilon+M_N$. Now choose $M=\max(M_1,\dots,M_{N-1},M_N+\epsilon)$.

Now $|g_nf|\le M|f|$, so you can use DCT.