Let $f: [a,b] \rightarrow \mathbb R $ be a Riemann integrable function. Prove that $ \lim_{c \rightarrow b-} \int_a^cf =\int_a^bf$.

Question

Let $f: [a,b] \to \mathbb{R}$ be a Riemann integrable function. Prove that

$$\lim_{c \to b^-} \int_a^cf =\int_a^bf.$$

It seems easier to look at this using the properties of integrals, and estimating $\left|\int_c^bf\right|$, rather than the definition of the integral. Any thoughts?

Answer 1

$f$ is Riemann integrable on $[a,b]$ implies that $f$ is bounded on $[a,b]$. (In general, it is the first result when we start the discussion the Riemann integral) Then, $|f(x)|\leq M$ for some $M>0$, and observe that for $a<c<b$, we have

$$ |\int_{a}^bf(x)dx-\int_{a}^{c}f(x)dx|=|\int_{c}^bf(x)dx|\leq \int_{c}^b |f(x)|dx\leq \int_{c}^b Mdx = M(b-c). $$ Finally, just letting $c\rightarrow b-$.

Answer 2

I propose another proof. By Fundamental Theorem of Calculus the function $$F(x)=\int_a^x f(t)\text{d}t$$ is continuous, whenever $f$ is Riemann-integrable. Hence $$\lim\limits_{c\to b^-} F(c)=F(b)$$ which is the statement we need.