Let $(X,\Sigma,\mu)$ be a measure space.


The semi-finite version of $\mu$ on $(X,\Sigma)$ is denoted $\mu\_{\rm sf}$ and given by
$$
\mu\_{\rm sf}(E) = \sup\{\mu(A) \mid A \subseteq E \text{ measurable, } \mu(A) \lt \infty\}, \quad \text{for}E \in \Sigma .
$$



> 
> Consider the map $i: L^2(\mu)\longrightarrow L^2(\mu\_{\rm sf}),\;\;f\longmapsto f$. Is $i$ bijective?
> 
> 
> 



Identity map between $L^2(\mu)$ and $L^2(\mu_{\rm sf})$
Yes, it is bijective.


p4sch has already shown the map is well-defined and injective (in fact, an isometry). I claim it is also surjective. (I assume throughout that $\mu$ is a positive measure.)


Note it is an exercise to show that $\newcommand{musf}{\mu\_{\rm sf}} \musf$ is indeed a countably additive measure.


**Lemma.** Suppose $E \in \Sigma$ with $\musf(E) < \infty$. Then there exists a measurable $B \subseteq E$ with $\mu(B) = \musf(E)$.


*Proof.* By definition of $\musf$, for each $k$ we may find a measurable $A \subseteq E$ with $\infty > \mu(A\_k) \ge \musf(E) - 1/k$. Set $B\_n = A\_1 \cup \dots \cup A\_n$. Then $B\_n \subseteq E$ and $\mu(B\_n) \ge \musf(E) - 1/n$. Moreover, since $\mu(B\_n) < \infty$, we have $\mu(B\_n) \le \musf(E)$ by the definition of $\musf$. Now set $B = \bigcup\_{n=1}^\infty B\_n$. Continuity from below shows $\mu(B) \le \musf(E)$ and monotonicity shows $\mu(B) \ge \musf(E)$. So $B$ is as desired. $\Box$


In particular we have $1\_B \in L^2(\mu)$. As noted by p2sch, whenever $\mu(B) < \infty$ we have $\mu(B) = \musf(B)$. Hence $\musf(B) = \musf(E)$, so that $\musf(E \setminus B) = 0$ and thus $1\_B = 1\_E$ $\musf$-a.e. Hence $1\_E$ (or, more properly, its equivalence class) is in the image of $i$. By linearity, every $\musf$-simple function is in the image, so the image is dense in $L^2(\musf)$. But we previously showed $i$ is an isometry, so the image is also closed.


