I'm asked to show the following:
"If $f\in \bf{R}$ $[a,b], $ then $ \displaystyle\int_{a}^{b} f = \lim_{c \to a^{+}} \int_{c}^{b} f$.
Here, $\bf{R}$ stands for the set of Riemann integrable functions over the set $[a,b]$.
I'm not sure where to really go with this except making use of the definitions of a Riemann integrable function;
$|U(P,f)-L(P,f)|<\epsilon$ for some partition P,
$|S(P,f)-\int_{a}^{b} f| < \epsilon$ provided that $||P|| <\delta$
I'm not sure how to work the $\displaystyle \lim_{c\to a^+}$ into the inequalities.
Any hints/tips would be appreciated! Thanks.
You are given that $f$ is Riemann integrable on $[a,b]$ which simplifies the proof. A more general proposition of this type is that $f$ is bounded on $[a,b]$ and integrable on $[c,b]$ for all $c > a$. In this case you would first show that $f$ is integrable on $[a,b]$ and the stated limit is true.
Since $f$ is Riemann integrable on $[a,b]$, it is bounded . There exists a number $M > 0$ such that $|f(x)| \leqslant M $ for all $x \in [a,b]$.
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,
$$\int_a^bf(x) \, dx = \\ \lim_{c \to a+} \int_a^bf(x) \, dx \\ = \lim_{c \to a+} \int_a^cf(x) \, dx + \lim_{c \to a+} \int_c^bf(x) \, dx \\ = \lim_{c \to a+} \int_c^bf(x) \, dx $$