$f$ is a decreasing function whose integral converges. Does $\lim_{x \to \infty}xf(x) = 0$? My finals are over and I cannot help but ruminate over this particular problem. Could anyone help prove this?

Suppose $f$ is a continuous decreasing function on $[0,\infty)$ and $\int_0^\infty f(t)\, \mathrm{d}t$ converges.  Prove that $$\lim_{x\to\infty} xf(x)=0.$$

 A: Although it's unnecessary for a solution, let's connect the elements of this question to familiar statistical concepts.
Because $f$ is continuous and decreasing on $[0,\infty)$ it behaves like some multiple of a survival function for a distribution.  We can turn it into one by ensuring none of its values exceed $1.$  Since the continuity and decreasing properties imply $f(0)$ is finite, and if $f(0)=0$ the conclusion follows immediately, let's suppose $f(0)\gt 0$ and for all $t\ge 0$ define
$$S(t) = \frac{1}{f(0)} f(t).$$
This makes $S$ the survival function of a continuous distribution, because we can extend it to a continuous non-increasing function on all real numbers by setting $S(x)=1$ for all $x\lt 0.$
In light of this, we ought to begin to wonder whether the continuity assumption on $f$ is really necessary: perhaps the result is true of any survival function.
Let's turn to the question by proving the contrapositive: if $\lim_{x\to\infty}xf(x)$ is nonzero or does not exist, that implies $\int_0^\infty f(t)\,\mathrm{d}t$ does not converge.
Now because all $x f(x)$ are non-negative when $x\gt 0,$ the failure of $x f(x)$ to have zero as a limit means there must be a positive threshold $\epsilon$ that is exceeded by $x f(x)$ at an unbounded sequence of positive points.  The unboundedness implies we may select a subsequence  $x_1 \lt x_2 \lt \ldots \lt x_n \lt \ldots$ for which $x_{n+1} \ge 2x_n$ for all $n;$ that is,
$$\frac{x_{n+1}-x_n}{x_{n+1}}\ge \frac{1}{2}.$$
For all of these points, because $x_n f(x_n)\ge \epsilon$ and $x_n\gt 0,$
$$f(x_n) \ge \frac{\epsilon}{x_n},\ n=1, 2, 3, \ldots$$
Moreover, on each interval $(x_n, x_{n+1}],$ $f(x) \ge f(x_{n+1})$ (because $f$ is not increasing).  Combining this with the preceding inequality gives
$$f(x) \ge f(x_{n+1}) \ge \frac{\epsilon}{x_{n+1}} \text{ when } x_n \lt x \le x_{n+1}.$$
Use this as a lower bound to estimate the integral, breaking it into integrals over the intervals between the $x_n$ (and just ignoring the initial interval from $0$ to $x_1$):
$$\begin{aligned}
\int_0^\infty f(t)\,\mathrm{d}t &= \int_0^{x_1} f(t)\,\mathrm{d}t + \sum_{n=1}^\infty \int_{x_n}^{x_{n+1}} f(t)\,\mathrm{d}t \\
&\ge \sum_{n=1}^\infty \int_{x_n}^{x_{n+1}} \frac{\epsilon}{x_{n+1}}\,\mathrm{d}t \\
&= \epsilon\sum_{n=1}^\infty \frac{x_{n+1}-x_{n}}{x_{n+1}}\\
& \ge \epsilon\sum_{n=1}^\infty \frac{1}{2},
\end{aligned}$$
which diverges. Therefore $\int_0^\infty f(t)\,\mathrm{d}t$ must diverge too.  This negation of the convergence assumption completes the proof.
Looking back over the assumptions needed for this analysis, we find that the continuity of $f$ was not needed anywhere: it sufficed that $f$ was nondecreasing and non-negative.  Consequently, this result holds for all survival functions.
