Problem
Let $f(x)\in L^{1}(\mathbb{R}^N)\cap L^{\infty}(\mathbb{R}^N)$ and $S_t=\left\{x\in\mathbb{R}^N: |x_1|\le t\right\}$ with $t>0$. Let $\phi(t)$ the integral function $$\phi(t)=\int_{\mathbb{R}^N\setminus S_t}|f(x)|^2dx$$ Find $$\lim_{t\to 0^{+}}\phi(t) \ \ \ \mbox{and} \ \ \ \lim_{t\to +\infty}\phi(t)$$
My approach
Since $f(x)\in L^{1}(\mathbb{R}^{N})\cap L^{\infty}(\mathbb{R}^N)$ then $f(x)\in L^{2}(\mathbb{R}^{N})$.
Infact, by definition of integral
$$\phi(t)=\int_{\mathbb{R}^N\setminus S_t}|f(x)|^2dx=\int_{\mathbb{R}^{N}}\chi_{\mathbb{R}^N\setminus S_{t}}(x)|f(x)|^2dx\le\\ \\ \le \int_{\mathbb{R}^{N}}|f(x)|^2dx\le \|f\|_{\infty}\int_{\mathbb{R}^{N}}|f(x)|dx=\|f\|_{\infty}\|f\|_{1}<+\infty$$
This means that $\phi(t)$ is a well posed and bounded function:
$$0\le \phi(t)\le \|f\|_{\infty}\|f\|_{1}\ \ \ \forall t\in (0,+\infty)$$
Note that for $0<t_1<t_2$, $S_{t_1}\subset S_{t_2} \implies\mathbb{R}^N\setminus S_{t_1}\supset\mathbb{R}^N\setminus S_{t_2}$ so:
$$\phi(t_1)=\int_{\mathbb{R}^{N}\setminus S_{t_1}}|f(x)|^2dx\ge \int_{\mathbb{R}^{N}\setminus S_{t_2}}|f(x)|^2dx=\phi(t_2)$$
Hence $\phi(t)$ is a bounded decreasing function, so $\lim_{t\to 0^{+}}\phi(t)$ and $\lim_{t\to +\infty}\phi(t)$ exist and are finite.
Now, I would like to use the sequential criterion for a limit of a function:
$$\lim_{t\to 0^{+}}\phi(t)=\lim_{n\to +\infty}\phi\left(\frac{1}{n}\right)=\lim_{n\to +\infty}\int_{\mathbb{R}^{N}\setminus S_{\frac{1}{n}}}|f(x)|^2dx=\\ \\ \\ =\lim_{n\to +\infty}\int_{\mathbb{R}^{N}}\chi_{\mathbb{R}^{N}\setminus S_{\frac{1}{n}}}(x)|f(x)|^2dx$$
Since $\chi_{\mathbb{R}^{N}\setminus S_{\frac{1}{n}}}(x)\le 1$ the sequence
$$g_{n}(x):=\chi_{\mathbb{R}^{N}\setminus S_{\frac{1}{n}}}(x)|f(x)|^2\le |f(x)|^2 \ \ \forall x\in\mathbb{R}^{N}$$
I can use the dominated convergence theorem
$$\lim_{n\to +\infty}\int_{\mathbb{R}^{N}}\chi_{\mathbb{R}^{N}\setminus S_{\frac{1}{n}}}(x)|f(x)|^2dx=\int_{\mathbb{R}^{N}}\lim_{n\to+\infty}\chi_{\mathbb{R}^{N}\setminus S_{\frac{1}{n}}}(x)|f(x)|^2dx=\int_{\mathbb{R}^{N}}|f(x)|^2dx=\|f\|_{2}^2$$
For the limit $\lim_{t\to +\infty}\phi(t)$ I use the sequence $t_{n}=n$ so
$$\lim_{t\to +\infty}\phi(t)=\lim_{n\to +\infty}\int_{\mathbb{R}^N}\chi_{\mathbb{R}^{N}\setminus S_{n}}(x)|f(x)|^2dx=\int_{\mathbb{R}^N}\overbrace{\lim_{n\to +\infty}\chi_{\mathbb{R}^{N}\setminus S_{n}}(x)}^{=0}|f(x)|^2dx=0$$
Is this approach ok? Thanks.