Find a non-negative function on [0,1] such that $t\cdot m(\{x:f(x) \geq t\}) \to 0$ that is not Lebesgue Integrable Problem:
Find a non-negative function $f$ on $[0,1]$ such that $$\lim_{t\to\infty} t\cdot m(\{x : f(x) \geq t\}) = 0,$$
but $f$ is not integrable, where $m$ is Lebesgue measure.
My Attempt:
Let $f(x) = \frac{\chi_{(0,1]}}{\sqrt{x}}$. Then, 
\begin{align*}
\lim_{t\to\infty} t\cdot m(\{x \in [0,1]: f(x) \geq t\}) &= \lim_{t\to\infty} t \cdot m(\{x\in (0,1]: 1/\sqrt{x} \geq t\})\\
&=\lim_{t\to\infty} t \cdot m(\{x\in (0,1]: x \leq (1/t^2)\})\\
&= \lim_{t\to\infty} t \cdot m((0, (1/t^2)))\\
&= \lim_{t\to\infty} \frac{t}{t^2} = 0.
\end{align*}
However, $f(x)$ is integrable over this interval. I have tried functions looking like $f(x) = 1/x^p$ but I cannot find any that will work here. There is a hint that says there is a monotonic function that fits this description.  
Also, does anyone know of a list of non-Lebesgue integrable functions on $[0,1]$? I feel as though I could use this for many counterexamples if one were to exist. Thanks!
 A: The example given in the previous answer was not quite accurate and in any case overcomplicated. Let $I_n$ denote the interval $(\delta_{n+1},\delta_n)$, where $\delta_n=2^{-n}/n$ and let
$$
f(x)=\sum_{n\geqslant 1}2^n\mathbf{1}_{I_n}(x).
$$
Then $f$ is non-negative, $2^km(f\geqslant 2^k)=\sum_{n\geqslant k}m(I_n)=\frac 1k$ and
$$
\int fdm=\sum_{n\geqslant 1 }2^n m(I_n)
$$
and $2^n m(I_n)$ behaves like $1/n$.
A: Just to show that the function $f(x)=\frac{1}{x|\log x|}\mathbb{1}_{(0,1)}(x)$ proposed by Davide Giraudo does not quite do the trick, but simple modifications (suggested by @Chival) of it will do.
The issue is the singularity at $1$.
Notice that the function $f(x)=\frac{1}{x|\log x|}$ is convex on $(0,1)$, attains a minimum value ($e$) at $t=e^{-1}$ and
$$\lim_{x\rightarrow0+}f(x)=\lim_{x\rightarrow1-}f(x)=\infty$$
For any $t>e$, there are exactly two points $0<a_t<e^{-1}<b_t<1$ such that $f(a_t)=f(b_t)=t$. In fact, $$a_t\xrightarrow{t\rightarrow\infty}0,\qquad b_t\xrightarrow{t\rightarrow\infty}1$$
by the convexity and monoticity of $f$ on $(0,e^{-1})$ and $(e^{-1},1)$. Then
$$t\,m(f>t)=t\,m\big((0,a_t)\cup(b_t,1)\big)=ta_t+t(1-b_t)\xrightarrow{t\rightarrow\infty}1$$
for
$$ta_t=\frac{1}{|\log a_t|}\xrightarrow{t\rightarrow\infty\infty}0,$$
and
$$t(1-b_t)=f(b_t)(1-b_t)=-\frac{1-b_t}{b_t\log b_t}\xrightarrow{t\rightarrow\infty}1.$$
The last bit follows by a simple application of L'Hospital theorem:
$$\frac{b-1}{b\log b}\sim \frac{1}{1+\log b}\xrightarrow{b\rightarrow1-}1$$

Notice that any truncation of $f$, say $g(x)=\frac{1}{x|\log x|}\mathbb{1}_{(0,e^{-1})}$, will do the job.
