Problems taking the limit in $\int_a^b f=\lim_{c\to a}\int_c^b f$ from definitions Let $f$ be bounded on $[a,b]$ and Riemann integrable for each $c$ with $a<c<b$.  I need to show that

$f$ is Riemann integrable on $[a,b]$, and $\int_a^b f=\lim_{c\to a}\int_c^b f$.

My problem is that I'm not sure how to go about taking the limit on the RHS.  I can start with $\epsilon > 0 $ as always, but then by the time I've gotten done definition chasing (taking an arbitrary partitions, looking at the sup and inf in the the upper and lower sums, etc) everything has become so confusing that I can't make any headway.  How do I unravel the definitions here to get something tractable?
Note: I am not trying to use any "big theorem" shortcuts here that just do it all in one step, I just want to prove it directly from the definitions, with minimal corner cutting. Also, because it sometimes seems relevant to mention this on this site, this is for self-study, not homework.
 A: As $f$ is bounded on $[a,b]$, we have $|f(x)| \leqslant M $ for all $x \in [a,b]$.
Choose $x_1$ such that
$$x_1 - a < \frac{\epsilon}{4M}.$$
As $f$ is integrable over $[x_1,b]$, for every $\epsilon > 0$ there is a partition $P': x_1 < x_2 < \ldots < x_n = b$ such that the difference between the upper and lower Darboux sums satisfies
$$U(P',f) - L(P',f) < \frac{\epsilon}{2}.$$
Consider the partition $P: a = x_0 < x_1 < x_2 < \ldots < x_n = b$ of $[a,b]$. Then the difference between the upper and lower sums is
$$U(P,f) - L(P,f) =  U(P',f) - L(P',f) + [\sup_{a \leqslant x \leqslant x_1}f(x) -  \inf_{a \leqslant x \leqslant x_1}f(x)](x_1-a) \\ <  \frac{\epsilon}{2} + 2M(x_1-a) \\ < \epsilon.$$
Therefore, $f$ satisifies the Riemann criterion  and is integrable over $[a,b]$.
Furthermore, $f$ is integrable over $[a,c]$ and
$$0 \leqslant \left|\int_a^cf(x) \, dx\right|\leqslant \int_a^c|f(x)| \, dx\leqslant M(c-a)\\ \implies \lim_{c \to a} \int_a^cf(x) \, dx =0.$$
Hence,
$$\int_a^bf(x) \, dx = \\ \lim_{c \to a} \int_a^bf(x) \, dx   \\ = \lim_{c \to a} \int_a^cf(x) \, dx  + \lim_{c \to a} \int_c^bf(x) \, dx  \\ = \lim_{c \to a} \int_c^bf(x) \, dx $$
