Question about integrating measures of pullbacks 
Let $f\in L^p([0,1])$ for some $1\leq p < \infty$. Define $h(t)=\mu \{ x \in [0,1]:|f(x)|>t \}$ for $0\leq t < \infty$. Prove that $h \in L^1([0,\infty)).$

My intuition is that $\int_{\mathbb R} h(t) dt \leq \int_{[0,1]} f(x) dx.$ The bounded space affords us that $h (t) < \infty$ for all $t.$ 
My initial thought at a solution for this problem is to utilize the old technique of demonstrating its truth for indicator functions, then for simple functions, and then taking limits. I am having trouble stating this clearly, however, and would appreciate any input. 
 A: You can do it using Fubini's Theorem:
$$ \int_{[0,1]} |f(x)| \, dx = \int_{[0,1]} \int_0^{|f(x)|} dy \, dx = \int_{[0,1]} \int_0^\infty I_{|f(x)|>y} \, dy \, dx \\ = \int_0^\infty \int_{[0,1]} I_{|f(x)|>y} \, dx \, dy = \int_0^\infty \mu(|f(x)|>y)\, dy = \int_0^\infty h(y) \, dy .$$
A: I'm sure you meant to write $h(t) = \mu \{ x \in [0,1] : |f(x)| > t \}$. You don't say much about the measure $\mu$ on $[0,1]$. Is it finite so that $L^p \subset L^1$ for $p > 1$? I suspect that's needed for this problem.
You specified $f \in L^{p}[0,1]$, which implies
$$
            t^ph(t) = t^{p}\mu \{ x \in [0,1] : |f(x)| > t \} \le \int_{|f| > t} |f(x)|^{p}d\mu(x) \le \|f\|_{p}^{p}.
$$
So $h$ is finite for $0 < t < \infty$, and $t^p h(t)$ tends to 0 as $t\uparrow\infty$ by the dominated convergence theorem. And $t^ph(t)$ remains bounded near $t=0$.
The function $h$ is a non-increasing function of $t$ which may or may not be infinite at $t=0$ because you don't say anything about $\mu$. Let $\mathcal{P}=\{a=x_{0} < x_{1} < \ldots < x_{n}=b\}$ be a partition of $[a,b] \subset (0,\infty)$. Then
$$
      \int_{0}^{1}\chi_{(a,b]}(|f(x)|)|f(x)|\,d\mu(x) 
          = \sum_{j=0}^{n-1}\int_{0}^{1}\chi_{(x_{j},x_{j+1}]}(|f(x)|)|f(x)|d\mu(x)
$$
yields
$$
        -\sum_{j=0}^{n-1}x_{j}^p\{ h(x_{j+1})-h(x_{j})\} \le \int_{a <|f|\le b}|f(x)|^pd\mu(x) \le -\sum_{j=0}^{n-1}x_{j+1}^p\{ h(x_{j+1})-h(x_{j})\}
$$
As the norm of the partition $\mathcal{P}$ tends to zero, the two sums converge to the same Riemann-Stieltjes integral of $-x^p$ with respect to $h(x)$, and, therefore,
$$
                    -\int_{a}^{b}x^pdh(x)=\int_{a < |f| \le b}|f(x)|^pd\mu(x).
$$
Integrating by parts,
$$
            -x^{p}h(x)|_{a}^{b}+\int_a^bpx^{p-1}h(x)dx=\int_{a < |f| \le b}|f(x)|^pd\mu(x).
$$
By earlier remarks, $x^ph(x)$ tends to 0 as $x\uparrow\infty$ and remains bounded as $x\downarrow 0$. Therefore,
$$
    \int_a^{\infty}px^{p-1}h(x)dx=\int_{a < |f|}|f(x)|^pd\mu(x)+a^ph(a).
$$
The right side remains bounded as $a\downarrow 0$, which means that the limit of the left side exists as $a \downarrow 0$, which then implies that $\lim_{a\downarrow 0}a^ph(a)$ exists. Finally,
$$
     \int_0^{\infty}px^{p-1}h(x)dx=\int_0^1 |f|^p d\mu+\lim_{a\downarrow 0}a^ph(a).
$$
If the measure $\mu$ is finite, then the above holds for $p=1$ because $L^{p}\subset L^{1}$ in that case. If not, then the last right on the right may not be finite for $p=1$, unless you start with $f\in L^{1}$. I'm not quite sure about that point.
