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(When I write integrable I mean Riemann-integrable)

Let $A \subseteq \mathbb{R}^m$ be a block, $f_n:A\rightarrow R$ be integrable functions and $f_n \rightarrow_{un} f$. Proof that $f$ is integrable and $\int_A f_n \rightarrow \int_A f$

I tried to use here the Lebesgue Criterion. (http://www.math.ncku.edu.tw/~rchen/Advanced%20Calculus/Lebesgue%20Criterion%20for%20Riemann%20Integrability.pdf)

Since each $f_i$ is integrable, the set of discontinuities of each $f_i$ has measure $0$. Since the convergence is uniform, the set of discontinuities of $f$ will also have measure $0$, then $f$ is integrable. My "proof" is just a sketch, I believe it's needing formalization, but I'm not sure how to do it. I'm also not sure how to analyse the sequence of the integrals either.

Can someone help me? Thanks.

Math1000
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    By integrable, you mean Lebesgue integrable or Riemann integrable? – Yiorgos S. Smyrlis May 27 '16 at 23:15
  • Riemann integrable. Edited. Thanks. –  May 27 '16 at 23:18
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    http://math.stackexchange.com/a/24175/192336 – Moya May 27 '16 at 23:18
  • if $\int_A 1 dx $ is finite then it is obvious, since $\int_A f_n(x) dx = \int_A f(x) dx + \int_A (f_n(x)-f(x) ) dx $ and $|f_n(x)-f(x)| < \epsilon$ for $n$ large enough. – reuns May 27 '16 at 23:20
  • I don't get your idea of "the set of discontinuities has measure $0$", you are on the wrong way. see $f_n(x) = \sum_{m=1}^n \frac{{ mx}}{m^2}$ where ${x} = x - \lfloor x \rfloor$. where is $f(x) = \lim_{n \to \infty} f_n(x)$ continuous ? – reuns May 27 '16 at 23:25
  • @user1952009 but wouldn't a suitable argument using this approach work for $m=1$? –  May 27 '16 at 23:27
  • @AhmedHussein : hmm ? no ok if the teacher wants to consider the measure of the set of discontinuities, do it.. but it seems weird to me – reuns May 27 '16 at 23:29
  • @user1952009 not that this was how I would approach the problem, but you are saying that you don't get the OP's idea. I am suggesting that a suitable argument works for $m=1$. For $m\ge2$, there isn't such a theorem to be applied at all –  May 27 '16 at 23:30
  • @AhmedHussein : do you mean the $m$ in $\mathbb{R}^m$ ? why would it be different when $m=2$ ? and no I don't get so well the idea behind the measure of the set of discontinuities, it seems that we need a complicated definition of the Riemann integral for it to be true (what about $g(x) = 1$ when $x$ is rational, $g(x) = 0$ otherwise ? It doesn't seem Riemann integrable to me, since $\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N g(n/N)$ is not the same as $\lim_{N \to \infty}\frac{1}{N}\sum_{n=1}^N g(n/N+1/N^\pi)$) – reuns May 27 '16 at 23:37
  • What exactly do you mean by a "block"? – Math1000 May 28 '16 at 03:05
  • @user1952009 yes, that's what I mean. I am referring to Lebesgue's characterization of Riemann integrable functions which only works for functions $[a,b] \to \Bbb R$. My idea was that if $A_i$ denotes the set of discontinuities of $f_i$, then the set of discontinuities of $f$ is a subset of $\cup_i A_i$ (which has measure zero) –  May 28 '16 at 10:14

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Remember that: $f$ is integrable on $A$ iff, for every $\varepsilon>0$, there exists a partition $P$ of $A$, such that $$ U(f,P)-L(f,P)<\varepsilon $$ where $L(f,P)$, $U(f,P)$ are the lower and upper sums of $f$ corresponding to $P$.

Let now $\varepsilon>0$. Since $f_n\to f$ uniformly on $A$, there exists an $N\in\mathbb N$, such that $n\ge N$, implies that $$ \sup_{x\in A}\lvert\, f_n(x)-f(x)\rvert<\frac{\varepsilon}{2(\mu(A)+1)}, $$ where $\mu(A)$ is the volume of $A$. Let $P$ a partition of $A$ for which $U(f_N,P)-U(f_N,P)<\frac{\varepsilon}{2}$. Note that, the partition $P$ divides $A$ in $K$ sub-blocks $\varDelta A_1,\ldots,\varDelta A_K$ of volumes $\mu(\varDelta A_1),\ldots,\mu(\varDelta A_K)$, respectively, and set $M_j(\,f_N),m_j(\,f_N)$ be the supremum and infimum of $f_N$ in $A_j$. Then $$ L(\,f_N,P)=\sum_{j=1}^K \mu(A_j)\,m_j(\,f_N), \quad U(\,f_N,P)=\sum_{j=1}^K \mu(A_j)\,M_j(\,f_N) $$ and $$ \lvert m_j(f_N)-m_j(f)\rvert < \frac{\varepsilon}{2(\mu(A)+1)},\,\,\, \lvert M_j(f_N)-M_j(f)\rvert < \frac{\varepsilon}{2(\mu(A)+1)} $$ and hence $$ U(f,P)-L(f,P)\le \cdots \le \frac{\varepsilon}{2}+\frac{\varepsilon \mu(A)}{2(\mu(A)+1)}<\varepsilon. $$ ETC

Yiorgos S. Smyrlis
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