Prove that improper integral $\int_{a}^{b} f(x) dx$ converges

Question

Here is the assignment.

Let $f: \ ]a,b]\rightarrow \mathbb{R}$ be non-negative and Riemann-integrable over $[c,b]$ for every $a<c<b$. Prove that improper integral $\int_{a}^{b} f(x) dx$ converges if and only if function $c\rightarrow\int_{c}^{b} f(x) dx$ is bounded.

I am able to prove that if improper integral $\int_{a}^{b} f(x) dx$ converges, then function $c\rightarrow\int_{c}^{b} f(x) dx$ is bounded:

$\int_{c}^{b} f(x) dx=\int_{a}^{b} f(x) dx-\int_{a}^{c} f(x) dx<\int_{a}^{b} f(x) dx\leq M$

However this other way around seems to be difficult. I can't even understand why is the statement true?

Answer 1

The function $$ c \mapsto \int_c^b f(x) \mathrm{dx}$$

is defined on $]a,b]$, and decreasing because $f$ is non-negative.

If you assume that this function is bounded, that implies that it has a finite limit when $c \rightarrow a$. By definition, it is equivalent to say that the improper integral $$\int_a^b f(x) \mathrm{dx}$$ converges.