The product of two nonnegative, improperly integrable functions is also improperly integrable.

Question

True or False: The product of two nonnegative, improperly integrable functions is also improperly integrable.

I was given both the problem and the proof that may or may not be true. I think the question is false but I couldn't come up with a counterexample that works or find the flaw in the proof. Does anyone have any suggestions for a counterexample of why the proof doesn't work?

Proof: Suppose $f$ and $g$ are two functions that are improperly integrable on $[c, \infty)$, where $c$ is any real number. Thus $\int_c^{\infty}f(x)dx\lt \infty$. We can conclude that $f(x) \to 0$ as $x \to \infty$. (If not, suppose there exists $\epsilon\gt 0$, s.t. $f(x) \gt \epsilon$ for large $x$. Then by the Comparison Theorem for Integrals, $\int_c^d f(x)dx\gt \epsilon(d-c)\to \infty$ as $d\to \infty$.) By definition of a limit, there exists $N \in \mathbb R$ s.t $0 \le f(x) \lt 1$ for $x > N$. This implies that $0 \le f(x)g(x) \lt g(x)$ for $x > N$. Since $g$ is improperly integrable on $[c, \infty)$, then by the Comparison Theorem for Improper Integrals, $fg$ is improperly integrable.

Answer 1

The conclussion that $f(x)\to0$ is wrong. $f$ may even be unbounded. Counterexample: $f\colon[1,\infty)\to[0,\infty)$ defined as equal to $n^2$ on $[n,n+1/n^4]$, and zero everywhere else. $f$ is nonnegative and integrable, but $f^2$ is not.

Answer 2

First of all, if $\int_0^\infty f(x)dx$ exists, you can not conclude that $\lim_{x\to\infty} f(x)$ exists, let alone equal to zero. There are several proofs of this fact across math.SE.

Second, a hint: take $a_n=(-1)^n n^{-1/2}$, then the sum $\sum_{n\ge 1}a_n$ converges, yet the sum $\sum_{n\ge 1}a_n^2$ does not.