Prove an equality ($L^P$ spaces) Prove the equality
$$\int |f(x)|^p dx=\int_0^\infty pt^{p-1}m(\left\lbrace x:|f(x)|\geq t\right\rbrace)dt$$
for $p\geq 1$.
My first idea was to try to prove this via induction.  For the case $p=1$, the left side becomes $\|f\|_1,$ and that's going to be finite since we are in $L^1$.  Now I think the right hand side becomes
$$t\cdot m(\left\lbrace x:|f(x)|\geq t\right\rbrace)\Big\vert_0^\infty=\lim_{t\rightarrow\infty}t\cdot m(\left\lbrace x:|f(x)|\geq t\right\rbrace).$$  By a previous exercise I have done a few weeks ago, this limit should go to zero.
As it stands, I can't get the two sides equal to each other, and that is only in the $p=1$ case.  Is there a better way to attack this?
 A: 
Lemma. Let $(X,M,\mu)$ be a $\sigma$-finite measurable space and $p\in[1,+\infty)$. Then for measurable non-negative function $f$ and measurable set $A\in M$ we have
  $$
\int\limits_A f(x)^p d\mu(x)=p\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt
$$
  where $F_{f,A}(t)=\mu(\{x\in A:f(x)>t\})$

Proof #1. Using Fubini theorem for positive functions we can say that
$$
\int\limits_A f(x)^p d\mu(x)=
\int\limits_A \int\limits_{(0,f(x))}pt^{p-1}dt d\mu(x)=
\int\limits_A \int\limits_{(0,+\infty)}pt^{p-1}\boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)dt d\mu(x)=
$$
$$
\int\limits_{(0,+\infty)}\int\limits_A pt^{p-1}\boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)d\mu(x) dt =
\int\limits_{(0,+\infty)}pt^{p-1}\int\limits_A \boldsymbol{1}_{\{x\in X:f(x)>t\}}(x)d\mu(x) dt =
$$
$$
\int\limits_{(0,+\infty)}pt^{p-1}\mu(\{x\in A:f(x)>t\}) dt =
\int\limits_{(0,+\infty)}pt^{p-1}F_{f,A}(t) dt
$$
Proof #2. Assume that $h(x)=\sum\limits_{k=1}^m c_k \boldsymbol{1}_{E_k}$ is a simple function. Without loss of generality we assume that numbers $\{c_k:k\in\{1,\ldots,m\}\}$ are positive and increasing and sets $E_1,\ldots,E_m$ are pairwise disjoint with positive measure. We easily see that
$$
h(x)^p=\sum\limits_{k=1}^m c_k^p \boldsymbol{1}_{E_k}
$$
$$
F_{h,A}(t)=
\begin{cases}
\sum\limits_{k=1}^{m} \mu(E_k\cap A)&\text{ if }\quad t\in(0,c_1)\\
\sum\limits_{k=1}^{m-1} \mu(E_k\cap A)&\text{ if }\quad t\in[c_1,c_2)\\
\ldots\ldots\ldots\ldots\ldots\ldots\\
\mu(E_m\cap A)&\text{ if }\quad t\in[c_{m-1},c_m)\\
0&\text{ if }\quad t\in[c_m,+\infty)
\end{cases}
$$
Taking formally $c_0=0$ we obtain
$$
p\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt=
p\sum\limits_{k=1}^m\sum\limits_{r=k}^m\mu(E_k\cap A)\int\limits_{[c_{k-1},c_k)}t^{p-1}dt=
$$
$$
\sum\limits_{k=1}^m\sum\limits_{r=k}^m\mu(E_k\cap A)(c_{k}^p-c_{k-1}^p)=
\sum\limits_{k=1}^m c_{k}^p\mu(E_k\cap A)=\int\limits_{A} h(x)^pd\mu(x)\tag{1}
$$
For general case consider increasing sequence of simple positive functions $\{h_n:n\in\mathbb{N}\}$ which pointwise converges to $f$. Then by Beppo-Levi theorem
$$
\int\limits_{A}f(x)^pd\mu(x)=\lim\limits_{n\to\infty}\int\limits_{A}h_n(x)^p d\mu(x)
$$
Since $\{h_n:n\in\mathbb{N}\}$ is a non-decreasing sequence, then does $F_{h_n,A}$. From theorem of continuity of measure for all $t\in(0,+\infty)$ we have $\lim\limits_{n\to\infty}F_{h_n(t),A}=F_{f,A}(t)$. Again appling Beppo-Levi we see
$$
\lim\limits_{n\to\infty}\int\limits_{(0,+\infty)}t^{p-1}F_{h_n,A}(t)dt=
\int\limits_{(0,+\infty)}t^{p-1}F_{f,A}(t)dt\tag{2}
$$
Result follows from $(1)$ and $(2)$.
