In our analysis lecture our professor talked about the following assumption to which he said, that he does not have a proof:
If for all sequences $(a_k)_{k\in \mathbb{N}},(b_k)_{k\in \mathbb{N}}$
with $a_k\to -\infty, b_k\to \infty$ already $\lim_{k\to \infty}
\int_{a_k}^{b_k} f(x)\,\mathrm{d}x=\alpha$ holds (meaning that the
limit is distinct and it exists), then the following improper integrals
exist as well: $\int_0^\infty f(x)\,\mathrm{d}x$ and $\int_{-\infty}^0
f(x)\mathrm{d}x$ and it is $\alpha=\int_{-\infty}^0
f(x)\mathrm{d}x+\int_0^\infty f(x)\,\mathrm{d}x$.
Now the problem:
We only presuppose the convergence of $\int_{a_k}^{b_k} f(x)\,\mathrm{d}x$
. In doing so, constant sequences $a_k$ or $b_k$ are excluded since $a_k\to -\infty$ and $b_k\to \infty$ have to hold.
How does the convergence of $ \int_{0}^{b_k} f(x)\,\mathrm{d}x$ and $\int_{a_k}^{0} f(x)\,\mathrm{d}x$ follow now? Needless to say, the $0$ can of course be replaced by another constant.
If we assume the existence of two sequences $a_k$ or $b_k$, for which that does not apply, then it's probably possible to do a proof by contradiction (if one skillfully considers subsequences). Most likely a lot of cases have to be distinguished, because one does not know why both part-integrals $\int_{0}^{b_k} f(x)\,\mathrm{d}x$ and $\int_{a_k}^{0} f(x)\,\mathrm{d}x$ not converge. They could be unbounded, or oscillating (or jump back and forth) or both.
Does someone here know how to prove the assumption or know where I can find such a proof on the internet? (Because my professor said that he doesn't have an obvious solution to this question)
Best Answer
Let $b_k, c_k \geq 0$ be sequences such that $\lim_{k\to\infty} b_k = \lim_{k\to\infty} c_k = \infty$ and suppose that $b_k < c_k$ for all $k \in \mathbb{N}$. By assumption,
$$\lim_{k\to\infty} \int_{b_k}^{c_k} f(x)~\mathrm dx = \lim_{k\to\infty}\left( \int_{-k}^{c_k} f(x)~\mathrm dx - \int_{-k}^{b_k} f(x)~\mathrm dx \right) = \alpha - \alpha = 0.$$
By the Cauchy criterion (cf here), this shows that $$\lim_{y\to\infty} \int_{0}^{y} f(x)~\mathrm d x $$ exists.