If the left Riemann sum of a function converges, is the function integrable? If the left Riemann sum of a function over uniform partition converges, is the function integrable?
To put the question more precisely, let me borrow a few definitions first. Pardon my use of potentially non-canon definitions of convergence. Given a bounded function $f:\left[a,b\right]\to\mathbb{R}$,


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*A partition $P$ is a set $\{x_i\}_{i=0}^{n}\subset\left[a,b\right]$ satisfying $a=x_0\leq x_1\leq\cdots\leq x_n=b$.

*The norm of a partition $\newcommand\norm[1]{\left\lVert#1\right\rVert}\norm{P}:=\max_{0\leq i\leq n}|x_i-x_{i-1}|$

*The left Riemann sum of $f$ over partition $P$ is $l(f,P):=\sum_{i=1}^nf(x_{i-1})(x_i-x_{i-1})$ 

*The left Riemann sum of $f$ is said to converge to $L$ iff $\newcommand\norm[1]{\left\lVert#1\right\rVert}\forall\epsilon>0, \exists\delta>0:\norm{P}<\delta$ implies $\left|l(f,P)-L\right|<\epsilon$

*A uniform partition $P_n$ of $n$ divisions is defined by $x_i=a+\frac{b-a}{n}i$

*The left Riemann sum of $f$ over uniform partitions is said to converge to $L$ iff $\forall\epsilon>0, \exists N\in\mathbb{N}: n\geq N\implies \left|l(f,P_n)-L\right|<\epsilon$


Now, is the following statement true?
If the left Riemann sum of $f$ converges to $L$, $f$ is Riemann integrable and its Riemann integral $\int_a^b f$ equals $L$.
In particular, I am curious whether the following limited case is true.
If the left Riemann sum of $f$ over uniform partitions converges to $L$, $f$ is Riemann integrable and its Riemann integral $\int_a^b f$ equals $L$.
My hunch is that the statements above are not true. But I can't come up with a counter example. Can someone give me some help here please?
 A: In this context "Cauchy integral" has the meaning you know.
It is a fact that if a function is bounded and Cauchy integrable over $[a,b]$, then it is also Riemann integrable over that interval.
It seems that there is no elementary proof of this theorem.
The proof in Kristensen, Poulsen, Reich A characterization of Riemann-Integrability, The American Mathematical Monthly, vol.69, No.6, pp. 498-505, (theorem 1), could be considered elementary because plays only with Riemann sums but is an indigestible game.
Note that there exist unbounded functions Cauchy integrable.
Also the use of regular partitions is enough to define Riemann integral.
See Jingcheng Tong Partitions of the interval in the definition of Riemann integral, Int. Journal of Math. Educ. in Sc. and Tech. 32 (2001), 788-793 (theorem 2).
I repeat that the use of only left (or right) Riemann sums with only regular partitions doesn't work.
A: No, one example is the function $$f(x) = \begin{cases} 1: & \; x \notin \mathbb{Q}; \\ 0: & x \in \mathbb{Q} \end{cases}$$ for which the left Riemann sum for any uniform partition of $[0,1]$ is always zero since $$f(a + (b-a)i/n) = 0$$for all integers $i$. This Riemann sum gives very little useful information about $f$ and in fact $f$ is not Riemann integrable at all.
