I'm struggling a bit with this problem I've been given. It regards Riemann integrability for a function $f$. It reads like this:
Decide whether the statement is true or false by giving a proof if it is true and a counterxample if it is false:
"Let $f:[a,b]\rightarrow \textbf{R}$ be a bounded function. Then $f$ is Riemann-integrable on $\textbf{R}$ if and only if $$L:=\lim_{n\to\infty}\sum_{k=1}^{n}f(x_{k}*)\frac{b-a}{n}\in\textbf{R},$$
where $f(x_{k}^{*})$ denotes the middle point of the interval $$\left[a+\frac{(k-1)(b-a)}{n},\ a+\frac{k(b-a)}{n}\right].$$ In this case, $L=\int_{a}^{b}f(x)\,dx$."
I've proven one direction:
($\Rightarrow$) Suppose $f$ is Riemann-integrable. Then $U(P,f)-L(P,f)<\epsilon$ for some partition P. Let this partition $P$ be $P=(a, a+\frac{b-a}{n}, ... , a+\frac{(n-1)(b-a)}{n}, b)$. Now, with this partition and with $x_{k}^*$ as the middle point of $[x_{k-1},x_{k}]$ we have:
$|\sum_{k=1}^{n}f(x_{k}^{*})\frac{b-a}{n}-\int_{b}^{a}f(x)\,dx|<\epsilon\Rightarrow\text{lim}_{n\to\infty}\sum_{k=1}^{n}f(x_{k}*)\frac{b-a}{n}=L\in\textbf{R}$.
Furthermore, $L=\int_{b}^{a}f(x)\,dx$.
Now it is the proof for the other direction that has me confused. I understand that I want to use that the given limit exists to aid me in proving that f is Riemann integrable and most of my attempts boil down to showing that:
$U(P,f)-L(P,f)=\sum_{k=1}^{n}(M_{k}-m_{k})\Delta_{x_{i}}\leq|\sum_{k=1}^{n}f(x_{k}*)\frac{b-a}{n}-L|<\epsilon$
Where $M_{k}$ is the supremum of $f$ in the interval $[x_{k-1}, x_{k}]$ and $m_{k}$ is the infimum.
Am I heading in the right direction? Is there something I haven't seen? Is the statement even true or does it break when trying to prove the converse? Would appreciate all kinds of hints, explanation or help possible.