Taking limit of a parameter that appears in the boundary of an improper integral

Question

I would like to take the limit in that expression :

$$A(\sigma) = \int^{f(\sigma)}_{- \infty} \Phi(y) dy $$

Here $\Phi$ is a $\mathcal C^{\infty} $ function, and $f(\sigma)$ is well behaved, with those two assumptions :

$$ \lim_{\sigma \to \infty} f(\sigma) = \infty$$

$$ \lim_{\sigma \to 0} f(\sigma) = \infty$$

How can you prove that $$ \lim_{\sigma \to \infty} A(\sigma) = \int^{ \infty}_{- \infty} \Phi(y) dy $$

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What I tried is using the fact that $\Phi$ is a continuous function so an anti derivative exists so (FTC) :

$$ A( \sigma ) = F(f (\sigma) ) - \lim_{x \to -\infty} F(x) $$

and I'd like to say that since $F$ is an anti derivative, it is continuous so I can take the limit of $\sigma$ inside.

What do you think?

Answer 1

If $\Phi$ is integrable yes (just apply DCT). Otherwise, it's false in general. A counter example would be $$\Phi(x)=\begin{cases}0&x\leq 0,\\\sin(x)&x>0,\end{cases}$$ and $f(\sigma )=\sigma $ (the limit at $0$ is irrelevant here).

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Edit

In measure theory, if you have a measure space $(X,\mathcal M,\mu)$, then if $E\subset X$ and $f:X\to \mathbb R$ is measurable,
$$\int_Ef\,\mathrm d \mu:=\int_X \boldsymbol 1_Ef\,\mathrm d \mu.$$

In your context, $$\int_{-\infty }^{f(\sigma )}\Phi(y)\,\mathrm d y:=\int_{\mathbb R}\boldsymbol 1_{(-\infty ,f(\sigma ))}(y)\Phi(y)\,\mathrm d y.$$ So, if you set $\psi_\sigma (y):=\boldsymbol 1_{(-\infty ,f(\sigma ))}(y)\Phi(y),$ then $$\int_{-\infty }^{f(\sigma )}\Phi(y)\,\mathrm d y:=\int_{\mathbb R}\psi_\sigma (y)\,\mathrm d y.$$

The aim is to prove when $$\lim_{\sigma \to \infty }\int_{-\infty }^{f(\sigma )}\Phi(y)\,\mathrm d y:=\lim_{\sigma \to \infty }\int_{\mathbb R}\psi_\sigma (y)\,\mathrm d y=\int_{\mathbb R}\Phi(y)\,\mathrm d y\tag{E}$$ holds. Several situations arises :

• If $\Phi\in L^1(\mathbb R)$, then (E) holds by DCT (since $|\psi_\sigma |\leq \Phi$ and $\psi_\sigma(y) \underset{\sigma \to \infty }{\longrightarrow } \Phi(y)$).

• If $\Phi\geq 0$ and mesurable, but $\Phi\notin L^1(\mathbb R)$, then (E) holds by MCT (since $(\psi_\sigma )_\sigma $ is positive and increasing).

• If $\int_{-\infty }^\infty \Phi(y)\,\mathrm d y$ is a convergent improper integral (or is $\pm \infty $), but $\Phi^+,\Phi^-\notin L^1(\mathbb R)$ (the positive and negative part of $\Phi$), then $$\lim_{\sigma \to \infty }\int_{-\infty }^{f(\sigma )}\Phi(y)\,\mathrm d y=\int_{-\infty }^\infty \Phi(y)\,\mathrm d y,$$ holds, by the fact that :

Proposition 1: If a function $g:\mathbb R\to \mathbb R$ has a limit at $+\infty $, i.e. if there is $\ell\in\mathbb R\cup\{\pm \infty \}$ s.t. $$\lim_{x\to \infty }g(x)=\ell,$$ then $$\lim_{x\to \infty }g(h(x))=g\left(\lim_{x\to \infty }h(x)\right)=\ell,$$ whenever $h:\mathbb R\to \mathbb R$ is such that $$\lim_{x\to \infty }h(x)=\infty .$$

So, taking $g(x)=\int_{-\infty }^x \Phi(y)\,\mathrm d y,$ (as far as $g(x)$ is well defined for all $x$) and apply the previous proposition will gives you the claim.

• Notice that the Proposition 1 can also be applied to the two first cases, i.e. when $\Phi\in L^1(\mathbb R)$ or $\Phi\geq 0$, measurable, but $\Phi\notin L^1(\mathbb R)$.

• Finally, if $\int_{-\infty }^\infty \Phi(y)\,\mathrm d y$ is not a convergent improper integral, then the result won't hold as shows the counter-example given at the very beginning.

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The term "not convergent improper integral" can be ambiguous. By it, I mean that the function $\Phi$ is Riemann integrable on all compacts intervals, but the limit $$\lim_{(x,z)\to(-\infty ,+\infty )}\int_x^z\Phi(y)\,\mathrm d y,$$ doesn't exist.