# Let $f_n$ uniformly convergent to$f$. Proof that $f$ is integrable and $\int_A f_n \rightarrow \int_A f$

(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.

• By integrable, you mean Lebesgue integrable or Riemann integrable? – Yiorgos S. Smyrlis May 27 '16 at 23:15
• Riemann integrable. Edited. Thanks. – user286485 May 27 '16 at 23:18
• 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

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