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Suppose a function $f$ is in both $L^1(\mathbb{R})$ and $L^2(\mathbb{R})$. For $m=1,2,\ldots$, let $$f_m(x) = f(x)\cdot \chi_{[-m,m]}(x)\cdot \chi_{\{ \lvert f(y)\rvert \leqslant m\}}(x)$$ where $\chi$ is the characteristic function. Then why is $f_m(x)\rightarrow f(x)$ in both $L^1$ and $L^2$?

This was used in an answer to this post, but I don't understand the explanation there. Perhaps someone can explain it in simpler terms, or show the proof?

PJ Miller
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    Use dominated convergence. $f_m \to f$ pointwise, and $|f_n - f|$ is dominated by $2|f|$, while $|f_n - f|^2$ is dominated by $(2|f|)^2$. By assumption both dominating functions are integrable. – Nate Eldredge Nov 11 '13 at 04:31

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As Nate Eldredge noted, since $|f_m|\rightarrow |f|$ pointwise, and $|f_m|$ is dominated by the integrable function $|f|$, and $|f_m|^2$ is dominated by the integrable function $|f|^2$, so the conclusion follows by the dominated convergence theorem.

PJ Miller
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