# Showing $\lbrace f\in L^2(\mathbb{R}):\int_{\mathbb{R}}\vert f \vert<\infty)\rbrace$ is a Borel Subset of $L^2(\mathbb{R})$

My idea is to attempt to write this set as a intersection of unions of sets which are Borel sets and then to use the fact the fact that the Borel sets form a $\sigma$-algebra. We have

$$\bigg\lbrace f\in L^2(\mathbb{R}):\int_{\mathbb{R}}\vert f \vert<\infty)\bigg\rbrace=\bigcup_{N\in \mathbb{N}}\bigg\lbrace f\in L^2(\mathbb{R}):\int_\mathbb{R}\vert f \vert <N\bigg\rbrace\\=\bigcap_{M\in \mathbb N}\bigcup_{N\in \mathbb{N}}\bigg\lbrace f\in L^2(\mathbb{R}):\int _{-M}^M\vert f\vert<N\bigg\rbrace$$

I am not sure how I can further expand in terms of unions and intersetions, but I don't see how the set $\lbrace f\in L^2(\mathbb{R}):\int_{-M}^M\vert f \vert <N\rbrace$ is a Borel set of $L^2(\mathbb{R}).$ It would suffice to have $\vert f\vert ^2$ in the integrand since that would imply our set is open and therefore Borel, but that isn't necessarily the case since not each Borel set is open. Otherwise, how should I attempt to describe this set if I'd like to conclude that it is a Borel set of $L^2(\mathbb{R})$?

For $M, N > 0$ consider

$$L^2(\mathbb{R}) \longrightarrow L^2(-M, M) \longrightarrow L^1(-M,M) \longrightarrow \mathbb{R}$$

The first arrow is the restriction, the second arrow is inclusion (see $L^p$ and $L^q$ space inclusion), the third arrow is the $L^1$ norm.

The composition of these three maps is $$f \mapsto \int_{-M}^M |f|$$ which is continuous, since it is the composition of continuous maps. Call $C_M^N = \{ f \in L^2 (\mathbb{R}) : \int_{-M}^M |f| < N\}$. Then these sets are clearly open subsets of $L^2 (\mathbb{R})$, since they are the counterimage of an open set via a continuous map.

Finally, as you said, this is enough to conclude.

• When you say "counterimage of an open set" which open set are you referring to? – The Substitute Mar 4 '15 at 1:28
• The open set $(-1, N)$. – Crostul Mar 4 '15 at 9:49
• Do you mean $(-N,N)$? – The Substitute Mar 4 '15 at 11:16
• It's the same, the norm is always $\geq 0$ – Crostul Mar 4 '15 at 11:40