$f(x)$ is positive, continuous, monotone and integrable in (0,1]. Is $\lim_{x \rightarrow 0} xf(x) = 0$? I'm having trouble with this question from an example test.
We have a positive function $f(x)$ that's monotone, continuous and integrable in $(0,1]$. Is $\lim_{x \rightarrow 0} xf(x) = 0$?
Progress
The only problematic case seems to be when $f(x)$ is unbounded and monotonic decreasing. For that case, I found out that $xf(x)=\int_{0}^{x} f(x)dt$ and that $0\leq xf(x)\leq \int_{0}^{x} f(t)dt$. From here I'm not sure how to go on.
Thanks!

If $f$ is improperly integrable over $[0,1]$, then $\lim\limits_{a\rightarrow0^+}\int_a^1 f(x)\,dx=L$ for some finite number $L$.
Note, for any $b$ in $(0,1)$:
$$\eqalign{
L =\int_0^1f(x)\,dx  = \int_0^bf(x)\,dx +\int_b^1f(x)\,dx\cr
}
$$
Now, letting $b\rightarrow0^+$, we have that $ \int_b^1f(x)\,dx$ converges to $L$, which implies that
$$
\lim_{b\rightarrow0^+}\int_0^b f(x)\,dx=0.
$$
