Is $f$ Riemann integrable?

Question

Let $f:[a,b] \rightarrow \mathbb{R}$ be a bounded function. Suppose that there exists a sequence $(P_{n})_{n}$ of partitions of $[a,b]$ so that $\lim_{n\rightarrow \infty} L(f,P_{n})=L= \lim_{n \rightarrow \infty} U(f,P_{n})$, where $L$ denotes the lower sum and $U$ the upper sum.

Does it follow that $f$ is Riemann integrable? (If it is, then I'm certain the value of the integral is equal to $L$.)

Answer 1

By assumption we know that, for every $\varepsilon>0$, there exist partitions $Q$ and $Q'$ of $[a,b]$ for which $$L(f,Q)>L-\frac12\varepsilon $$ and $$U(f,Q')<L+\frac 12\varepsilon$$ If $P$ is a common refinement of $Q$ and $Q'$, one has $$L(f,P) \ge L(f,Q)>L-\frac12\varepsilon $$ and $$U(f,P) \le U(f,Q')<L+\frac 12\varepsilon$$ Subtracting last inequalities, we obtain $$U(f,P)-L(f,P) < \varepsilon $$ It follows that $f$ is integrable on $[a,b]$ by a well known criterion.

Since $$L(f,P_n)\le \int_a^b f \le U(f,P_n)$$ the statement is proved.

Answer 2

Yes.

Given $\epsilon>0$, we have to show that there exists $\delta>0$ such that for every partition $P$ finer than $\delta$, we have $L-\epsilon < L(f,P)$ and $U(f,P)<L+\epsilon$.

Let $M$ be a bounf for $f$, $|f(x)|<M$ for all $x\in[a,b]$. From what is given, we know that there exists $N$ such that $L-\frac\epsilon2<L(f,P_N)\le U(f,P_N)<L+\frac\epsilon2$. Let $x_0=a<x_1<\ldots < x_m=b$ be the points making up partition $P_N$. For some $\delta>0$, consider a partition $P$ that is finer than $\delta$. If $f_{P_n}$ and $f_P$ are the respective staircase functions used to compute the lower sum with respect to $P_N$ and $P$, then we have $f_P(x)\ge f_{P_N}(x)$ for all $x$ except possibly those in the up to $m+1$ intervals of $P$ that contain a point of $P_N$. For these intervals we of course have the trival estimate $f_P(x)\ge -M$ and $f_{P_N}(x)\le M$. Therefore $$L(f,P)+2M(m+1)\delta \ge L(f,P_N).$$ Hence it suffices to make $\delta<\frac\epsilon{4M(m+1)}$ in order to ensure $$L(f,P)\ge L(f,P_N)-\frac\epsilon2>L-\epsilon.$$ A similar argument shows that then also $$U(f,P)\le U(f,P_N)+\frac\epsilon2<L+\epsilon.$$