Sufficient condition for convergence of integral

Question

Given a function $f\colon \mathbb R \to \mathbb R$ and that $$ \lim_{t\to \infty}\int_{-t}^tf(x) \ dx=1, $$ does the integral $$ \int_{-\infty}^{\infty}f(x) \ dx $$ necessarily converge? I've been trying to come up with a counterexample rather than a proof, but to no avail yet.

Answer 1

Take any integrable function $g$ with compact support such that $\int_{-\infty}^\infty g(x) dx = 1$. Then take some function integrable on compact intervals with $h(-x) = -h(x)$ for all $x \in \mathbb{R}$ but which is not integrable. Define $f := h + g$. For large $t > 0$ it holds $\int_{-t}^t f(x) dx = 1$, so $\lim_{t\to \infty}\int_{-t}^t f(x) dx = 1$, but since $h$ is not integrable and $g$ is, $f$ is not integrable. So, for example $g(x) = 1$ for $|x|\leq 1/2$ and $g(x) = 0$ for $|x|>1/2$ and $h(x) = x$.