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

$$0 \leqslant \left|\int_a^cf(x) \, dx\right|\leqslant \int_a^c|f(x)| \, dx\leqslant M(c-a)\\ \implies \lim_{c \to a+} \int_a^cf(x) \, dx =0.$$


$$\int_a^bf(x) \, dx = \\ \lim_{c \to a+} \int_a^bf(x) \, dx \\ = \lim_{c \to a+} \int_a^cf(x) \, dx + \lim_{c \to a+} \int_c^bf(x) \, dx \\ = \lim_{c \to a+} \int_c^bf(x) \, dx $$

  • $\begingroup$ Thanks so much! I never thought to split it up into two intervals and then use the standard estimate/squeeze theorem to show that one of the integrals was 0. $\endgroup$ – Sorey Mar 3 '16 at 22:14
  • $\begingroup$ @Sorey: You're welcome. You would only need to consider sums to prove the more general proposition above. $\endgroup$ – RRL Mar 3 '16 at 22:21

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