Let $f : [a, \infty) \to \mathbb R$ be an integrable function on each $[a, r], r \gt a$. Show that $$\lim_{r \to +\infty} \int_a^x f(t) dt$$ exists if, and only if, for every $\epsilon \gt 0$, there exists $A \gt a$ such that $$M \lt A \lt B \to \mid \int_A^B f(t)dt \mid \lt \epsilon$$
My attempt
($\Rightarrow$)
Suppose the integral converges to $L$, let $\epsilon \gt 0$, using the definition of convergence, we can take $M \ge a$ sufficiently large such that if $A \ge M$ we have
$$\mid \int_a^A f(x)dx - L \mid \lt \frac{\epsilon}{2} $$
taking $B \gt A \ge M$ we have
$$\mid \int_A^B f(x)dx - L\mid=\mid \int_a^B f(x)dx - \int_a^A f(x)dx \mid = \mid \int_a^B f(x )dx - L + L - \int_a^A f(x)dx \mid \le$$
by triangular inequality
$$\mid \int_a^B f(x)dx - L \mid + \mid \int_a^A f(x)dx - L \mid \le \frac{\epsilon}{2} +\frac{\epsilon }{2}=\epsilon$$
so is worth
$$\mid\int_A^B f(x)dx - L\mid \lt \epsilon$$ as we wanted.
($\Leftarrow$)
For natural n, we define
$$a_n = \int_a^n f(x)dx$$ For $\epsilon \gt 0$
there is $m \ge a$ such that if $m, n \ge M$ we have
$$\mid a_n - a_m\mid = \mid\int_m^nf(x)dx\mid \lt \epsilon$$
Hence $(a_n)$ is a Cauchy sequence, so it is convergent, let $\lim a_n = L$. Given $\epsilon \gt 0$ we choose $M \ge a $($M$ natural) such that $n \gt m$, and any $A_1,B_1$ with $B_1 \gt A_1 \gt M$ we have $\mid a_n - L \mid \lt \frac{\epsilon}{2}$
$$\mid\int_{A_1}^{B_1} f(x)dx\mid \lt \frac{\epsilon}{2}$$
Taking $A \ge M +1$ then $\left \lfloor{A}\right \rfloor \gt M$ such that
$$\mid\int_a^A f(x)dx - L\mid = \mid\int_a^{\left \lfloor{A}\right \rfloor} f(x)dx - L + \int_\left \lfloor{A}\right \rfloor^A f(x)dx\mid \le \mid a_{\left \lfloor{A}\right \rfloor} - L\mid +\mid\int_{\left \lfloor{A}\right \rfloor}^A f(x)dx\mid \lt \frac{\epsilon}{2} + \frac{\epsilon}{2}= \epsilon$$
The proof is good enough ?
Thanks in advance for any help.
It looks mostly good but you are a little inconsistent in notation. The question as you post it says 'exists $A>a$ such that for all $y>x>A$', but then instead of $y,x$ you use $A,B$, so you change notation and $A$ changes it's meaning. Also missing + sign in the last calculation, before last inequality.