# Cauchy-type criterion for uniform convergence of improper integral

Suppose $$f=f(x,t)$$ is defined on the region $$D:=A \times [c,\infty)\subseteq \mathbb R^2,$$ and suppose $$\int_c^{\infty} f(x,t)dt$$ exists for all $$x \in A.$$ Call this improper integral uniformly Cauchy if, for all $$\epsilon >0,$$ there exists $$M>c$$ such that $$\left \lvert \int_u^v f(x,t)dt \right \rvert< \epsilon$$ whenever $$u,v \geq M$$ and $$x \in A.$$

I have been able to show that uniformly convergent $$\implies$$ uniformly Cauchy, and I am wondering if the converse is true. I think I was able to show that it is, but I am not entirely convinced by my proof. For brevity I will omit the details, but essentially I tried to show that if the improper integral is uniformly Cauchy, then the sequence of functions defined by $$F_n(x):=\int_c^{c+n} f(x,t)dt$$ is uniformly Cauchy, and therefore converges uniformly to $$F(x):=\lim_{n\to \infty} F_n(x).$$ Then I think I was able to show that $$\int_c^{\infty} f(x,t)dt$$ converges uniformly to $$F.$$

So essentially my question is, is it even true that uniformly Cauchy $$\implies$$ uniformly convergent? If so, then does my proof sketch seem correct? Thanks!

• The converse is true. You have the right idea missing the details.
– RRL
Jul 15 '20 at 0:57

Having shown that $$\{F_n(x)\}_n$$ is uniformly Cauchy, we can claim there exists a function $$I:A \to \mathbb{R}$$ such that $$F_n(x) \to I(x)$$ uniformly on $$A$$.

We have,

$$\left|\int_c^d f(x,t) \, dt - I(x) \right| \leqslant \underbrace{|F_n(x) - I(x)|}_{\alpha(x)} + \underbrace{\left|\int_c^d f(x,t) \, dt - \int_c^{c+n} f(x,t) \, dt\right|}_{\beta(x)}$$

There exists a positive integer $$N(\epsilon)$$ such that if $$n \geqslant N(\epsilon)$$, then $$\alpha(x)< \epsilon/2$$ for all $$x \in A$$.

Finally, using the uniform Cauchy condition, show that there exists $$C(\epsilon)> c +N(\epsilon)$$ (independent of $$x$$) such that if $$n = N(\epsilon)$$ and $$d > C(\epsilon)$$, then

$$\beta(x) = \left|\int_{c+N(\epsilon)}^d f(x,t) \, dt\right|< \epsilon/2,$$

for all $$x \in A$$.