Proof that $f_{n} = \chi_{\Omega_{n}}f \to f$ in $L^{2}$

Question

Let $f: \mathbb{R} \to \mathbb{C}$ be an unbounded measurable function and let $\mu$ be a finite measure on the Borel $\sigma$-algebra $\mathbb{B}(\mathbb{R})$ such that: $$\int_{\mathbb{R}}|f(x)|^{2}d\mu(x) < +\infty$$ For each $n \in \mathbb{N}$, let $\Omega_{n} := \{x \in \mathbb{R}: |f(x)| \le n\}$ and $\chi_{\Omega_{n}}$ its characteristic function, i.e. $\chi_{\Omega_{n}}(x) = 1 $ if $x \in \Omega_{n}$ and zero otherwise. I'm trying to prove that the sequence of functions $f_{n} := \chi_{\Omega_{n}}f$ is Cauchy and converges in $L^{2}(\mathbb{R},\mu)$ to $f$. My work so far is as follows: $$||f_{n}-f_{m}||^{2} = \int_{\mathbb{R}}|\chi_{\Omega_{n}}f-\chi_{\Omega_{m}}f|^{2}d\mu = \int_{\mathbb{R}}|\chi_{\Omega_{n}}-\chi_{\Omega_{m}}|^{2}|f|^{2}d\mu$$ Now, assuming $m \le n$, we notice that $\Omega_{m}\subseteq \Omega_{n}$, so that $|\chi_{\Omega_{n}}-\chi_{\Omega_{m}}| = \chi_{\Omega_{n}}-\chi_{\Omega_{m}} = \chi_{\Omega_{n}\setminus \Omega_{m}}$. Hence, $$||f_{n}-f_{m}||^{2} = \int_{\mathbb{R}}\chi_{\Omega_{n}\setminus\Omega_{m}}|f|^{2}d\mu = \int_{\Omega_{n}\setminus\Omega_{m}}|f|^{2}d\mu$$ But I'm stuck here. How can I prove this is Cauchy?

Answer 1

You probably want to use the dominated convergence theorem extended to $L^2$. Naturally, $|f_n(x)|^2 \le |f(x)|^2$ for every $n \in \mathbb{N}$ and $x \in \mathbb{R}$ so from the dominated convergence theorem, $$\underset{n\rightarrow \infty}{\lim}\int |f-f_n|^2d\mu = 0.$$

For more details for how to move from the $L^1$ version to any $L^p$ version of the theorem you can check https://en.wikipedia.org/wiki/Dominated_convergence_theorem in the Dominated convergence in Lp-spaces section.