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Rudin RCA p.32 exercise 10

Let $(X,\mathfrak{M},\mu)$ be a measure space and $\mu(X)<\infty$ and $\{f_n\}$ be a sequence of bounded complex measurable functions on $X$, and $f_n\rightarrow f$ uniformly on $X$. Prove that $\lim_{n\to\infty}\int_X f_n d\mu =\int_X f d\mu$.

Is 'boundedness' essential here? Here's my argument below and please tell me where am i misunderstanding..

Fix $\epsilon>0$. Then, there exists $N\in\mathbb{N} \text{ such that } n≧N \Rightarrow |f_n - f|< \epsilon$.

Let then a sequence $\{|f_n - f|\}_{n≧N}$ is dominated by $\epsilon$ and ,by assumtion, $\int_X \epsilon d\mu <\infty$. Thus, by Lebesgue's dominated theorem, $\lim_{n\to \infty} \int_X |f_n - f| d\mu = \int_X \lim_{n\to\infty} |f_n-f| d\mu = 0$. Q.E.D.

Thank you in advance

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  • $\begingroup$ How do you know that the integral exists otherwise? $\endgroup$
    – SBF
    Jan 15, 2013 at 13:08
  • $\begingroup$ @Ilya Oh right.. So the fake part is the last part. That is, [$|\int_X f_n-f d\mu|=0 \text{ implies } \int_X f_n d\mu - \int_X f d\mu =0$] is false, since it might me $\infty-\infty$. Am I right? $\endgroup$
    – Katlus
    Jan 15, 2013 at 13:15
  • $\begingroup$ Yes, of course if you know that $\|f-f_n\| = \sup_x |f(x)-f_n(x)|$ is less than $\epsilon$ then the difference $f-f_n$ is bounded and hence integrable. But it does not mean that $f$ and $f_n$ are integrable by themselves. $\endgroup$
    – SBF
    Jan 15, 2013 at 13:17
  • $\begingroup$ @Ilya Now i got it thank you! $\endgroup$
    – Katlus
    Jan 15, 2013 at 13:18
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    $\begingroup$ It seems a bit odd to use the dominated convergence theorem here . You can (and in fact did) estimate directly: if $n$ is so large that $\lVert f - f_n\rVert_\infty \leq \varepsilon$ then $\left\lvert \int f - \int f_n \right\rvert \leq \varepsilon \mu(X)$. All you need is that the difference of integrals makes sense and this is guaranteed by assuming integrability. $\endgroup$
    – Martin
    Jan 15, 2013 at 13:20

2 Answers 2

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The word bounded in the statement can be replaced by integrable. It includes a larger class of functions.

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  • $\begingroup$ I want to make it clear. Is "$f$ is Lebesgue Integrable" same as saying "$f\in L^1(\mu)$"? $\endgroup$
    – Katlus
    Jan 15, 2013 at 13:26
  • $\begingroup$ Yes, it's the definition (at least, the one I know). $\endgroup$
    – Davide Giraudo
    Jan 15, 2013 at 13:26
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In order to use the LDC theorem, you must know that $\,\int\limits_Xf_nd\mu\,$ exists, and from here that boundeness in this case means $\,\{f_n\}\,$ are integrable, so

$$\int\limits_Xf_nd\mu\xrightarrow[n\to\infty]{}\int\limits_Xfd\mu$$

by LDC theorem as you correctly showed.

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  • $\begingroup$ In OP LDC was applied to $f-f_n$ which is integrable. $\endgroup$
    – SBF
    Jan 15, 2013 at 13:19
  • $\begingroup$ Yes, I know...but in order to be ble to apply LDC to begin with it must be known $\,\int f_n\,$ exists $\endgroup$
    – DonAntonio
    Jan 15, 2013 at 13:20

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