I'm considering a problem about time varying domain in Chen Wenxiong and Li Congming 's study on $-\Delta u=\exp u$ in $\mathbb R^2$ and $\int_{\mathbb R^2} \exp u(x) \, d x< +\infty$.
LEMMA 1.1 (Ding). If $u$ is a solution of $-\Delta u=\exp u$ in $\mathbb R^2$ and $\int_{\mathbb R^2} \exp u(x) \, d x<$ $+\infty$, then $\int_{\mathbb R^2} \exp u(x) \, d x \geqslant 8 \pi$.
Proof. For $-\infty<t<\infty$ let $\Omega_t=\{x \mid u(x)>t\}$; one can obtain
$$
\begin{aligned}
\int_{\Omega_t} \exp u(x) \, d x & =-\int_{\Omega_t} \Delta u=\int_{\partial \Omega_t}|\nabla u| \,d s \\
-\frac{d}{d t}\left|\Omega_t\right| & =\int_{\partial \Omega_t} \frac{d s}{|\nabla u|} .
\end{aligned}
$$
By the Schwarz inequality and the isoperimetric inequality,
$$
\int_{\partial \Omega_t} \frac{d s}{|\nabla u|} \cdot \int_{\partial \Omega_t}|\nabla u| \geqslant\left|\partial \Omega_t\right|^2 \geqslant 4 \pi\left|\Omega_t\right| .
$$
Hence
$$
-\left(\frac{d}{d t}\left|\Omega_t\right|\right) \cdot \int_{\Omega_t} \exp u(x) \, d x \geqslant 4 \pi\left|\Omega_t\right|
$$
and so
$$
\frac{d}{d t}\left(\int_{\Omega_t} \exp u(x)\, d x\right)^2=2 \exp t \cdot\left(\frac{d}{d t}\left|\Omega_t\right|\right) \cdot \int_{\Omega_t} \exp u(x)\, d x \leqslant-8 \pi \exp t \cdot\left|\Omega_t\right| .
$$
Integrating from $-\infty$ to $\infty$ gives
$$\tag{1}
-\left(\int_{\mathbb R^2} \exp u(x) \,d x\right)^2 \leqslant-8 \pi \int_{\mathbb R^2} \exp u(x) \,d x
$$
which implies $\int_{\mathbb R^2} \exp u(x) \,d x \geqslant 8 \pi$ as desired.
The proof details before (1) is presented by the discussions between me and other users in Differentiation in time varying domain, I'm confused about the last step, how they integrate from $-\infty$ to $\infty$?
In detail, for the LHS, how to compute the limit of $\lim_{t \rightarrow +\infty} \int_{\Omega_t} \exp u(x) \, d x$ and $\lim_{t \rightarrow -\infty} \int_{\Omega_t} \exp u(x)\, d x$, I think $\lim_{t \rightarrow -\infty} \int_{\Omega_t} \exp u(x)\, d x$ must be $\int_{\mathbb R^2} \exp u(x) \,d x$ but I don't know how to deal with $\lim_{t \rightarrow +\infty} \int_{\Omega_t} \exp u(x) \,d x$, from the fact that $\lim_{t \rightarrow -\infty} \int_{\Omega_t} \exp u(x) \,d x = \int_{\mathbb R^2} \exp u(x) \,d x$ which is bounded and $\frac{d}{d t}\left(\int_{\Omega_t} \exp u(x) \,d x\right)^2=2 \exp t \cdot\left(\frac{d}{d t}\left|\Omega_t\right|\right) \cdot \int_{\Omega_t} \exp u(x) \,d x \leqslant 0 $ we know that when $t \rightarrow +\infty$, $\lim_{t \rightarrow +\infty} \int_{\Omega_t} \exp u(x) \, d x$ is also bounded, but based on the proof, it seems that it is 0.
For the RHS, how to integrate $\exp t \cdot\left|\Omega_t\right|$ from $-\infty$ to $\infty$? I try to use Reynolds transport theorem
\begin{align}
& \frac{\mathrm{d}}{\mathrm{d} t} \int_{\Omega(t)} \mathbb{A}(\boldsymbol{x}, t)\, \mathrm{d}^n \boldsymbol{x} \\[8pt]
= {} & \int_{\partial \Omega(t)}(\boldsymbol{V}(\boldsymbol{x}, t) \cdot \boldsymbol{n}(\boldsymbol{x}, t)) \mathbb{A}(\boldsymbol{x}, t) \,\mathrm{d}^{n-1} \boldsymbol{x}+\int_{\Omega(t)} \partial_t \mathbb{A}(\boldsymbol{x}, t) \,\mathrm{d}^n \boldsymbol{x}
\end{align}
I set $A_1=e^u$ and $A_2=e^t$, because on $\partial \Omega(t)$ we have $u=t$ so we have
$$\exp t \cdot\left|\Omega_t\right|=
\int_{\Omega t} e^t \,d x =\frac{d}{d t} \int_{\Omega t} e^t \,d x – \frac{d}{d t}\int_{\Omega t} e^u \,d x
$$
Now we have to face another question which is $\lim_{t \rightarrow +\infty} \exp t \cdot\left|\Omega_t\right|$ and $\lim_{t \rightarrow -\infty} \exp t \cdot\left|\Omega_t\right|$
Best Answer
I think your question is just basic real analysis, not at the research level, you should ask this question on stack exchange instead of overflow.
Since $\int_{\mathbb{R}^2} e^u \, dx <\infty$, for every $\epsilon>0$, there exists $\delta>0$ such that if $|A|<\delta$, then $\int_A e^u \, dx <\epsilon$. By Chebyshev's Inequality, it is easy to see that $|\Omega_t|\to 0$ as $t\to +\infty$. So $\lim_{t\to 0} \int_{\Omega_t} e^u \, dx=0$. On the other hand, by the description of the $L^1$ norm in terms of the distribution function, we have $$\int_{\mathbb{R}^2} e^u \, dx =\int_0^\infty |\Omega_{\ln s}| \, ds =\int_{-\infty}^{+\infty}e^t|\Omega_{t}| \, dt. $$ In the last step, we use the formula of change of variables by setting $s=e^t.$