Bounding integrals on a `moving' domain I've been reading through Schoen-Yau's proof of the positive mass theorem and find myself stuck on a particular estimate they prove. It seems to be simple multivariable calculus, but I haven't been able to prove it. Here's a simplified setup:
Let $S\subset\mathbb{R}^3$ be a non-compact surface and consider the set of exhautions $S_{\sigma}=S\cap B_\sigma(0)$, where $B_\sigma(0)$ is the ball of radius $\sigma$ with centre $0$. Clearly, $S_{\sigma_1}\subset S_{\sigma_2}$ if $\sigma_1<\sigma_2$ and $\{S_\sigma\}$ exhausts $S$ as $\sigma\rightarrow\infty$. They use the following argument to bound the integral,
$$
\int_S \frac{1}{1+r^a}=\int_{S_{\sigma}}\frac{1}{1+r^a}+\int_{\sigma_0}^\infty\left(\frac{d}{dt}\int_{S_t}\frac{1}{1+r^a}\right)dt\leq Area(S_{\sigma_0})+\int_{\sigma_0}^\infty\frac{1}{1+t^a}\left(\frac{d}{dt}Area(S_t)\right)dt.
$$
Here $a>2$ and $\sigma_0>0$ are constants. The first equality is just an application of the fundamental theorem of calculus. The $Area(S_\sigma)$ is clear. The term that is bothering me is the second term involving $\frac{d}{dt}Area(S_t)$. I'm not sure how they got that bound. Are they using some kind of Leibniz rule type formula?
 A: I don't know much about Leibniz' integral rule but this is some tricky computations, for which the sequence $(S_t)$ being increasing is fundamental. Write:
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t}  \int_{S_t} \frac{1}{1+r^a}\mathrm{d}r = \lim_{h\to 0} \frac{1}{h}\left(\int_{S_{t+h}}\frac{1}{1+r^a}\mathrm{d}r - \int_{S_t}\frac{1}{1+r^a}\mathrm{d}r \right)
\end{align}
Note that as $(S_t)$ is an increasing sequence of subsets:
$$
\int_{S_{t+h}}\frac{1}{1+r^a}\mathrm{d}r - \int_{S_t}\frac{1}{1+r^a}\mathrm{d}r = \int_{S_{t+h}\setminus S_t}\frac{1}{1+r^a}\mathrm{d}r.
$$
Now, you can bound the RHS by:
$$
\int_{S_{t+h}\setminus S_t} \frac{1}{1+r^a}\mathrm{d}r \leqslant \sup_{r\in S_{t+h}\setminus S_t} \left\{\frac{1}{1+r^a} \right\} \int_{S_{t+h}\setminus S_t} 1 \mathrm{d}r \leqslant\frac{1}{1+t^a}  \times \mathrm{Area}\left(S_{t+h} \setminus S_t \right),  
$$
because $t\leqslant r \leqslant t+h$ and $r \mapsto \frac{1}{1+r^a}$ is decreasing. Hopefully, this last term is, still because the sequence of sets is increasing, computable:
$$
\mathrm{Area}\left(S_{t+h} \setminus S_t \right) = \mathrm{Area}(S_{t+h}) - \mathrm{Area}(S_t).
$$
Finally, dividing by $h$ and taking $h\to 0$ gives the desire result.
