Let $\left\{f_{n}\right\}$ be a sequence of bounded functions that
converges uniformly to $f$ on the closed, bounded interval $[a, b] .$
If each $f_{n}$ is Riemann integrable over $[a, b]$, show that $f$
also is Riemann integrable over $[a, b]$. Is it true that $$ \lim _{n
\rightarrow \infty} (R)\int_{a}^{b} f_{n}=(R)\int_{a}^{b} f ? $$
This could be solved using upper and lower Darboux sums (as answered in other problems), but I was wondering if I could take an easier approach.
First, $f_n$ Riemann integrable implies that they are Lebesgue integrable.
Second, $\lim _{n
\rightarrow \infty} (R)\int_{a}^{b} f_{n}=\lim _{n
\rightarrow \infty} \int_{a}^{b} f_{n}=\int_{a}^{b} f$.
Lastly, to show that $\int_{a}^{b} f = (R)\int_{a}^{b} f$, it is enough to show that the set of discontinuities of $f$ is measure zero, by Lebesgue's theorem. Again, by Lebesgue's theorem, the set of discontinuities of each $f_n$ is measure zero. Can I use this and uniform convergence to show that $f$ is Riemann integrable?
Best Answer
If you are already aware of Lebesgue's criteria for Riemann integrability ($f$ is Riemann integrable iff $f$ is continuous everywhere outside a set of measure zero) then of course, you can deduce the integrability of $f$ from this. Suppose $D_n$ is the set of discontinuities of $f_n$ and let $D=\bigcup_nD_n$. Since $f_n$ converges uniformly to $f$, then $f$ is continuous on $[a,b]\setminus D$ (continuity is presented by uniform convergence) The countable union of sets of measure $0$ is also of measure $0$. Then $f$ is continuous outside $D$ (a set of measure $0$) and hence, $f$ is Riemann integrable.
If you are not aware of Lebesgue's result, you can prove integrability of $f$ through Darboux criteria: Here is a sketch of how one may approach this:
Given $\varepsilon>0$, there is $N_\varepsilon\in\mathbb{N}$ such that $\|f-f_n\|_u<\varepsilon/3(b-a)$ for all $n\geq N_\varepsilon$.
Since $f_N$ is Riemann integrable, there is a partition $\mathcal{P}_\varepsilon=\{a=x_0<\ldots<b_n=x_n\}$ of $[a,b]$ such that $U(f_N,P)-L(f_N,P)<\varepsilon/3$
It follows that
$\sup_{x\in[x_j,x_{j+1}]}f(x)=\sup_{x\in[x_j,x_{j+1}]}\big(f(x)-f_N(x)+f_N(x)\big)<\frac{\varepsilon}{3}(b-a) + \sup_{x\in[x_j,x_{j+1}]}f_N(x)$
$\inf_{x\in[x_j,x_{j+1}]}f(t)=\inf_{x\in[x_j,x_{j+1}]}\big(f(x)-f_N(x)+f_N(x)\big)> -\frac{\varepsilon}{3}(b-a) + \inf_{x\in[x_j,x_{j+1}]}f_N(x)$
Then $$\begin{align} U(f,P_\varepsilon)&=\sum^n_{j=1}\big(\sup_{x\in[x_j,x_{j+1}]}f(x)\big)\,(x_j-x_{j-1})\\ &<\frac{\varepsilon}{3}+\sum^n_{j=1}\big(\sup_{x\in[x_j,x_{j+1}]}f_N(x)\big)\,(x_j-x_{j-1})=\frac{\varepsilon}{3}+U(f_N,P_\varepsilon) \end{align}$$ and $$\begin{align} L(f,P_\varepsilon)&=\sum^n_{j=1}\big(\inf_{x\in[x_j,x_{j+1}]}f(x)\big)\,(x_j-x_{j-1})\\ &>\frac{\varepsilon}{3} +\sum^n_{j=1}\big(\inf_{x\in[x_j,x_{j+1}]}f_N(x)\big)\,(x_j-x_{j-1})=-\frac{\varepsilon}{3}+L(f_N,P_\varepsilon) \end{align}$$
Hence $$ U(f,P_\varepsilon) - L(f,P_\varepsilon)<\frac23\varepsilon +U(f_N,P_\varepsilon)-L(f,P_\varepsilon)<\varepsilon $$
This shows that $f$ is Riemann integrable.
To conclude, notice that since $|f(x)-f_n(x)|<\frac{\varepsilon}{3}(b-a)$ for all $x\in[a,b]$ and all $n\geq N_\varepsilon$ $$ \Big|\int^b_a(f-f_n)\Big|\leq\int^b_a|f-f_n|\leq \varepsilon/3,\qquad n\geq N_\varepsilon $$
This shows that $\lim_n\int^b_af_n=\int^b_af$.