how to prove the integral converges Suppose $$I=\lim_{x\rightarrow\infty}\lim_{b\searrow0}\int_{b}^{x}{g(y)dy}$$ exists and is finite, where $g$ is a continuous function from $\mathbb{R}^{+}$ to $\mathbb{R}$. Prove $$\lim_{x\rightarrow\infty,b\searrow 0}{\int_{b}^{x}{g(y)dy}}$$ exists and equals $I$.
Can someone give me hints? I do not know where to start it.
 A: EDIT: Usually one likes to see "why" something is true intuitively and then translate that understanding into a proof. Here I just followed the definitions and found I was done - never did understand "why" it was so. Until now: It's a special case of the following obvious fact:
Obvious Fact If $\lim_{x\to a}\left(\lim_{y\to b}(F(x)+G(y))\right)=I$ then $\lim_{(x,y)\to(a,b)}(F(x)+G(y))=I$.

Original:
Inserting some parentheses to make things absolutely clear: You're assuming that
$$I=\lim_{x\rightarrow\infty}\left(\lim_{b\searrow0}\int_{b}^{x}{g(y)dy}\right),$$
and you want to show that $$\lim_{(x,b)\to(\infty,0^+)}{\int_{b}^{x}{g(y)dy}}=I,$$right?
My first reaction was to say this does not follow. The corresponding implication with a function $F(x,b)$ in place of $\int_b^x g$ is not true. But of course, as I realized quickly when I started to construct a counterexample,  $\int_b^x g$ cannot be an arbitrary function of $x$ and $b$.
Suppose $I=0$ for simplicity and assume $\epsilon>0$. There exists $A$ so that
$$\left|\lim_{b\to0}\int_b^xg\right|<\epsilon\quad(x\ge A).$$It follows that
$$\left|\int_x^yg\right|<2\epsilon\quad(A\le x\le y).$$(Given $x,y\ge A$, choose $b$ so that $|\int_b^x|<\epsilon$ and $|\int_b^y|<\epsilon$.)
Now choose $\delta>0$ so that $$\left|\int_b^Ag\right|<\epsilon\quad(0<b<\delta).$$(There exists such a $\delta$ because the above implies that $|\lim_{b\to0}\int_b^A|<\epsilon$.)
If $x\ge A$ and $0<b<\delta$ then $$\left|\int_b^xg\right|\le\left|\int_b^Ag\right|+\left|\int_A^xg\right|<\epsilon+2\epsilon=3\epsilon.$$
