Suppose $f\in L^p[0,1]$ for some $p$ with $1\le p\le\infty$. Define $$h(t):=m(\{x\in[0,1]:|f(x)|>t\})$$ where $m$ is the Lebesgue measure. Show that $\int_0^\infty h(t)\; dt<\infty$.
My first thought was to try to use Chebyshev's inequality for finite $p$ as follows: $$ \begin{split} \int_0^\infty h(t)\; dt &= \int_0^\infty m(\{x\in[0,1]:|f(x)|>t\})\; dt \\ &\le \int_0^\infty\frac1{t^p} \int_{\{x\in[0,1]:|f(x)|>t\}}|f(x)|^p\; dmdt \end{split} $$ And then use the fact that $f$ is in $L^p$ to bound the inner integral by some number $M_t$ dependent on $t$ which should go to 0 as $t$ goes to infinity. But this doesn't seem to be the way to go as it isn't clear how to integrate the resulting $\frac{M_t}{t^p}$ and even if I could, it's not obvious to me whether the result would be finite. In fact, it seems that it won't be unless $M_t\to0$ as $t\to0$.
So what other methods should I try at this point?