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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.

81235
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    Your two limits of integrals appear to be them same. – Marc Dec 14 '15 at 16:28
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    I guess the OP asks to prove that taking the limit for $(x,b)\to(\infty,0^+)$ jointly is the same as taking it first wrt $b$ and then wrt $x$. – AndreasT Dec 14 '15 at 16:31

1 Answers1

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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 $$\left|\int_b^xg\right|\le\left|\int_b^Ag\right|+\left|\int_A^xg\right|<\epsilon+2\epsilon=3\epsilon.$$

David C. Ullrich
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  • So for my problem, I can define $f(y)=g(y)-\frac{I}{x-b}$? – 81235 Dec 15 '15 at 01:13
  • @81235 I have no idea what you're talking about, sorry... – David C. Ullrich Dec 15 '15 at 05:25
  • My question is what if $I\neq 0$ – 81235 Dec 15 '15 at 05:27
  • Another question is how you have $|\int_{b}^{x}|<\varepsilon$? From $|\lim_{b\rightarrow 0}\int_{b}^{x}{g}|<\varepsilon$? It is kind of weird that the integral $|\int_{x}^{y}{g}|$ can be very small if $x$ and $y$ goes to infinity. – 81235 Dec 15 '15 at 05:30
  • Let $h$ be any function with $\int h=I$ and such that $h$ vanishes on $(0,1)$ and vanishes on $(2,\infty)$. The proof I gave applies to $g-h$ in place of $g$, and then the conclusion for $g-h$ implies the conclusion for $g$. – David C. Ullrich Dec 15 '15 at 05:35
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    Hint: If $|\lim_{b\to0}\int_b^x|<\epsilon$ then $|\int_b^x|<\epsilon$ if $b$ is small enough. Similarly for $y$. Now if $b$ is small enough that both of those integrals are less than $\epsilon$ in absolute value their difference is less than $2\epsilon$. – David C. Ullrich Dec 15 '15 at 05:41