Proof of Cauchy's Criterion I would like to prove the following:
Cauchy's Criterion: If $f$ is locally integrable on $[a,b)$, then $\int_a^b f(x) \,dx $ exists $\iff \forall \epsilon >0$, $\exists r \in (a,b)$ such that $x_1,x_2 \in [r,b) \implies |\int_{x_1}^{x_2} f(x) \,dx| < \epsilon $. $(*)$
Here is what I have:
Define $F(x)  = \int_a^x f(t) dt$. Then by definition of  the improper integral, we can rewrite the statement as $\lim_{x \to b^-} F(x)$ exists $\iff \forall \epsilon >0$, $\exists r \in (a,b)$ such that $x_1,x_2 \in [r,b) \implies |F(x_2) - F(x_1)| < \epsilon$. $(1)$
Now I have already proven before that $\lim_{x \to b^-} F(x)$ exists $\iff \forall \epsilon >0$, $\exists \delta > 0$ such that $x_1,x_2 \in (b - \delta, b) \implies |F(x_2) - F(x_1)| < \epsilon$. $(2)$
So I would like to show that $(2) \implies (1) $, So that $(2) \implies (*)$.
Suppose $(2)$ holds. Fix $\epsilon > 0$. $\exists \delta > 0$ with $\delta < b - a$. Choose $r \in (b-\delta,b)$. Then $(2)$ holds for any $x_1,x_2 \in [r,b)$. However notice that $ a < b - \delta < r < b$ so $r \in (a,b)$. Thus $(2) \implies (1)$.
Is my ending argument valid? I have seen arguments where a bound is given on $\delta$ before, but I am not sure if this can always be done or only in specific circumstances.
 A: I'm just going to rewrite your assertions: Let $f$ be locally integrable in $[a,b)$ and
$F(x)=\int_a^xf(t)dt$ for $a\leq x<b$. It should be easy for you to show that the following assertions are equivalent:
$(i)$ $\forall\varepsilon>0$, $\exists\delta>0$ s.t. $\forall x_1,x_2\in[b-\delta,b)$, $|\int_{x_1}^{x_2}f(t)dt|<\varepsilon$;
$(ii)$ $\forall\varepsilon>0$, $\exists\delta>0$ s.t. $\forall x_1,x_2\in[b-\delta,b)$, $|F(x_2)-F(x_1)|<\varepsilon$.
Now, consider the following assertion:
$(*)$ The limit $\lim_{x\rightarrow b^-}\int_a^xf(t)dt=\lim_{x\rightarrow b^-}F(x)=:F(b^-)$ exists.
Let's show that $(*)\iff (ii)$
$(\Rightarrow)$
Suppose that $F(b^-)$ exists. Then, by definition of left lateral limit, given $\varepsilon>0$, there exists $\delta>0$ s.t. for every $x\in[b-\delta,b)$, we have $|F(x)-F(b^-)|<\varepsilon/2$. So, if $x_1,x_2\in [b-\delta,b)$, we have
$$|F(x_2)-F(x_1)|\leq |F(x_2)-F(b^-)|+|F(b^-)-F(x_1)|<\varepsilon,$$
so $(ii)$ holds.
$(\Leftarrow)$
Now, suppose $(ii)$ holds. Let $y_n=b-(b-a)/n$ for $n=1,2,\ldots$. Notice that $y_n\rightarrow b$, and that $(ii)$ directly implies that $\left\{F(y_n)\right\}_n$ is a Cauchy sequence. By the Cauchy criterion for $\mathbb{R}$, we have that $\left\{F(y_n)\right\}_n$ converges, say, for some $L$.
Let's show that $L=\lim_{x\rightarrow b^-}F(x)$. Given $\varepsilon>0$, take, by $(ii)$, an $\delta>0$ s.t. for every $x_1,x_2\in[b-\delta,b)$, we have $|F(x_1)-F(x_2)|<\varepsilon/2$.
Since $y_n\rightarrow b$ and $F(y_n)\rightarrow L$, there exists $N_0$ such that $y_{N_0}\in[b-\delta,b)$ and $|F(y_{N_0})-L|<\varepsilon/2$ (if you're not convinced about it, try to prove this. It's easy). Now, given any $x\in [b-\delta,b)$,
$$|F(x)-L|\leq |F(x)-F(y_{N_0})|-|F(y_{N_0})-L|<\varepsilon.$$
Therefore, $L=F(b^-)$ exists, so $(*)$ is valid.
