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

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?

• You don't want to estimate $\left|\int_a^b f\right|$, but, rather, $\left|\int_c^b f\right|$. – Ted Shifrin Apr 22 '17 at 17:09
• @TedShifrin that is what I meant to write, my mistake. I've made the appropriate edits. How would you go about estimating it though? – km24 Apr 22 '17 at 17:13
• Is $f$ bounded? – Ted Shifrin Apr 22 '17 at 17:14
• @TedShifrin the exercise does not specifically state whether or not $f$ is bounded – km24 Apr 22 '17 at 17:19
• Look up the definition of a Riemann integrable function. – Ted Shifrin Apr 22 '17 at 17:20

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.
$f$ 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 $c\rightarrow b-$.
• Putting here $x,y$ instead of $b,c$ we get by your argument (without any changes) that the function $F(x)$ defined in my answer is Lipschitz-continuous, so, it is uniformly continuous. – szw1710 Apr 22 '17 at 18:06