Identity map between $L^2(\mu)$ and $L^2(\mu_{\rm sf})$

Question

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?

Answer 1

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.

Answer 2

If $\mu$ is not a positive measure, we cannot say anything about the negative part $\mu^-$ with the help of $\mu_{sf}$: If $\mu$ is a signed-measure, use the Hahn-decomposition to conclude that $\mu_{sf}(E) = \mu_{sf}^+(E)$. Thus, the statement is already false, if there exist a measurable functions $f$ such that $$\int |f|^2 \mathop{d \mu^-} = \infty, \ \ \ \text{but} \ \ \ \ \int |f|^2 \mathop{d \mu^+} < \infty.$$ Assuming that $\mu$ is a positive measure, the statement is true.

In my first answer I have ignored that we only talk about equivalence classes of functions. We may have functions $f \neq g$ $\mu$-almost everywhere with $f \in L^2(\mu)$ and $g \notin L^2(\mu)$, but in this case $f = g$ $\mu_{sf}$-almost everywhere, as we will see below.

One simple example is $\Omega = \{1\}$, $\Sigma = P(\Omega)$, $\mu = \infty \cdot \delta_1$. Here $1_\Omega \notin L^2(\mu)$, but $\mu_{sf} =0$ and thus $1_\Omega =0$ $\mu_{sf}$-almost everywhere.)

We know that any integrable function is already located on a $\sigma$-finite measurable subset (that follows from the Chebyshev's inequality). Note that for any set $E \in \Sigma$ with finite measure, i.e. $\mu(E) < \infty$, we have $\mu_{sf}(E) = \mu(E)$, just by definition and monotonicity. Moreover, if $E$ is $\sigma$-finite (i.e. $E= \bigcup_{n=1}^\infty A_n$ with measurable $A_n$ satisfying $\mu(A_n) <\infty$), then also $\mu(E) = \mu_{sf}(E)$. (Proof: Taking $E_m = \bigcup_{n=1}^m A_n$ and using measure continuity, we get $\mu(E) \leq \mu_{sf}(E)$. On the other hand, we have for any measurable $A \subset E$ already $\mu(A) \leq \mu(E)$ by measure-monotonicity.)

That observation implies for any $f \in L^2(\mu)$ $$\int |f|^2 \mathop{d \mu} = 2 \int_0^\infty x \, \mu(|f| > x) \mathop{dx} = 2 \int_0^\infty x \, \mu_{sf}(|f| > x) \mathop{dx} = \int |f|^2 \mathop{d \mu_{sf}},$$ i.e. the map is well-defined (in the sense that for $f=g$ $\mu$-almost everywhere, we have also $f=g$ $\mu_{sf}$-almost everywhere) and injective.

This map is also surjective.

Proof: First, let us add a proof of the fact that $\mu_{sf}$ is a measure.

1. Of course, $\mu_{sf}(\emptyset)=0$.
2. Now, let $(A_n)_{n \in \mathbb{N}} \subset \Sigma$ be disjoint sets. If $\mu_{sf}(A_n) = \infty$ for some $n \in \mathbb{N}$, then we see easily that also $\mu_{sf}(E) = \infty$ for $E:= \bigcup_{n=1}^\infty A_n$. Thus, we can suppose that $\mu_{sf}(A_n) <\infty$ for all $n \in \mathbb{N}$. Taking $B_n \subset A_n$ with $\mu_{sf}(A_n) \leq \mu(B_n) + \frac{\varepsilon}{2^n}$ shows that $$\sum_{k=1}^n \mu_{sf}(A_k) \leq \sum_{k=1}^n (\mu(B_k)+\frac{\varepsilon}{2^k}) \leq \mu(\cup_{k=1}^n B_k) + \varepsilon \leq \mu_{sf}(E) + \varepsilon.$$ Thus $\sum_{k=1}^\infty \mu_{sf}(A_k) \leq \mu_{sf}(E)$. On the other hand, we find for any $A \subset E$ with finite measure that $\mu(A) = \sum_{n=1}^\infty \mu(A \cap A_n) \leq \sum_{k=1}^\infty \mu_{sf}(A_k).$

In order to prove that this map is surjective, first assume that $g = 1_A \in L^2(\mu_{sf})$, i.e. $\mu_{sf}(A) <\infty$. In this case, we can take $A_n \subset A$ with w.l.o.g. $A_n \uparrow$, $\mu(A_n) < \infty$ and $\mu(A_n) \uparrow \mu_{sf}(A)$. Define $B= \bigcup_{n=1}^\infty A_n$, then $\mu(B) = \mu_{sf}(A)$. (Note that it can happen that $\mu(A) = \infty$ as in the previous example.) Because of $\mu(B) = \mu_{sf}(B)$, we get $\mu_{sf}(A \setminus B) =0$ and thus $F(1_B)=1_A$, where $F \colon L^2(\mu) \rightarrow L^2(\mu_{sf})$ denotes the identity map.

Now any $g \in L^2(\mu_{sf})$ can be approximated by simple functions $g_n$. Moreover, by the previous step, we find simple functions $f_n \in L^2(\mu)$ with $F(f_n) =g_n$. Using that $F$ is an isometry, we see that $(f_n)_{n \in \mathbb{N}}$ is a Cauchy-sequence in $L^2(\mu)$, say, with limes $f \in L^2(\mu)$. One can check easily that $F(f) =g$.