$f(x)$ 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!
 A: 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
$$
\lim_{b\rightarrow0^+}\int_0^b f(x)\,dx=0.
$$
A: If $f$ is increasing, then $\lim_{x \to 0^+} f(x) = C < +\infty$ exists, so that the result is trivial. 
If $f$ is decreasing, $c < x$ implies $f(c) \ge f(x)$. Now by the mean value theorem for integrals, for every $0 < x < 1$, there exists $0 < c(x) < x$ such that 
$$
\int_0^x f(t) \, dt = f(c(x))x. 
$$
This gives
$$
0 \le x f(x) \le x f(c(x)) = \int_0^x f(t) \, dt \to 0.
$$
Hope that helps,
A: If there is some $\beta > 0$ and $L>0$ so that $xf(x)  \ge L$ on $(0,\beta)$, then 
$f(x) \ge L/x$ on $(0,\beta)$  and hence
$$
\int_0^\beta f(x) dx
$$
doesn't exist.
Otherwise for every $L$ there is a decreasing sequence $\{u_{L,n}\} \to 0$ such that
$u_{L,n}f(u_{L,n}) < L$. Hence by monotonicity,  $xf(x) < L$ on $(0,u_{L,1})$.
Now let $L\to 0$. It follows that $xf(x) \to 0$.
A: Just a longer comment - I'll show that improper integrals are continuous.
Improper integral on $(0,1]$ is just the limit of regular integrals on $[\delta,1]$. If the limit $L$ exists, then for $\delta < \delta_0$ this integral is $\varepsilon$-close to L. From the triangle inequality we know, that two integrals on $[\delta,1]$ and $[\delta',1]$ are $2\varepsilon$-close (note that $\delta,\delta'$ might also be zero here, then we get just $\varepsilon$). It means that when we integrate on an interval included in $[0,\delta_0]$, then the result is smaller than $2\varepsilon$. Moreover on $[\delta_0,1]$ the function is bounded by some $M$ (from its integrability). So, if we take an interval $[a,b]$ shorter than $\min(\delta_0,\varepsilon/M)$, then (by triangle inequality) the integral is smaller than $2\varepsilon + \varepsilon=3\varepsilon$, which shows the continuity of an improper integral.
A: As you pointed out $0\leq x f(x) \leq \int_{0}^{x} f(t)\, dt$. But $F(x):= \int_{0}^{x}f(t)\, dt$ is continuous, so $\lim_{x\to 0} F(x) = F(0) = 0$, and the result follows from the squeeze theorem.
