# Measurability of closed opeartor

Given separable Banach spaces $$(E, \Vert \cdot \Vert_E)$$ and $$(F, \Vert \cdot \Vert_F)$$. The Banach space $$E$$ is endowed with sigma algebra $$\mathcal{F}$$ which is generated by the open set of it. Similarly, let $$\mathcal{G}$$ be the sigma algebra of space $$F$$. Say $$A: D(A) \to F$$ is a closed operator where $$D(A)$$ is a subspace of $$E$$ and we assume $$D(A) \in \mathcal{F}.$$ My question is that is the map $$A: (D(A), \mathcal{F}|_{D(A)}) \to (F, \mathcal{G})$$ necessarily measurable? I know that when $$D(A)$$ is closed this is true, since via closed graph theorem, $$A$$ is in fact continuous. But in general, I don't know how to prove it.

In the separable case, it follows from the following standard result from descriptive set theory:

Theorem. Let $$X,Y$$ be Polish spaces, let $$f : X \to Y$$ be a Borel function, and let $$B \subset X$$ be a Borel set. If the restriction of $$f$$ to $$B$$ is one-to-one, then $$f(B)$$ is a Borel subset of $$Y$$, and the restriction of $$f^{-1}$$ to $$f(B)$$ is a Borel function.

See for instance Proposition 4.5.1 of Srivastava, A Course on Borel Sets.

Now to apply this theorem, let $$X = E \oplus F$$, let $$Y = E$$, and let $$f = \pi_E : E \oplus F \to E$$ be the projection onto $$E$$, which is continuous and in particular Borel. Let $$B$$ be the graph of $$A$$, which by assumption is closed and in particular Borel. Note $$\pi_E(B) = D(A)$$. Since $$B$$ is a graph, $$\pi_{E}|_B$$ is one-to-one, so by the theorem above, $$D(A)$$ is Borel and the restriction of $$\pi_E^{-1}$$ to $$D(A)$$, which is simply the map $$x \mapsto (x, Ax)$$, is Borel. Then $$A$$ is just the composition of this map with the continuous map $$\pi_F$$.

• @Nate Eldredge I see, thanks a lot for the help! – lye012 May 4 at 1:41

The answer is yes if you assume that $$E$$ and $$F$$ are separable. In the non-separable case I suspect it is false but someone else can write a separate answer for that.

To prove this (in the separable case), it suffices to prove that the functional $$f(x):= \|Ax\|_F$$ is measurable as a map from $$(D(A),\mathcal F|_{D(A)}) \to (\Bbb R_+,$$ Borel$$)$$. Indeed, if we can show that $$f$$ is measurable then the claim can be obtained as follows. Denote by $$B_F(z,r)$$ the ball of radius $$r>0$$ around $$z\in F$$. Note that for each $$y \in D(A)$$, the preimage of $$B_F(Ay,r)$$ under $$A$$ is just a translate of the set $$f^{-1}[0,r)$$ by the vector $$y$$, hence it is measurable. Next we take an open set $$U$$ in $$F$$. For each $$x \in A^{-1}(U)$$, choose $$r_x>0$$ such that $$B_F(Ax,r_x) \subset U$$. Then let $$U' = \bigcup_{x\in A^{-1}(U)} B_F(Ax,r_x)$$, and notice that $$A^{-1}(U) = A^{-1}(U')$$. Now since $$F$$ is separable, the open cover $$\{B_F(Ax,r_x)\}_{x\in U}$$ admits a countable subcover $$\{B_F(Ax_i,r_{x_i})\}_{i \in \Bbb N}$$. Then $$A^{-1}(U) = A^{-1}(U') = \bigcup_{i \in \Bbb N} A^{-1}(B_F(Ax_i,r_{x_i}))$$, which is measurable.

We will denote the graph norm by $$\|x\|_G := \|x\|_E+\|Ax\|_F$$, which makes $$D(A)$$ into a Banach space. To prove that $$f$$ is measurable, we can write it as the supremum of a countable collection of continuous functions. First notice by this other question that $$D(A)$$ is separable with respect to the graph norm (because $$E,F$$ are both separable) so let $$\{x_1,x_2,...\}$$ be a dense subset of its unit sphere. Use Hahn-Banach to obtain bounded linear functionals $$f_n:E \to \Bbb R$$ with the property that $$f_n(x_n) = 1$$ and $$\|f_n\|_{E \to \Bbb R} = \|x_n\|_E^{-1}$$ and also $$\|f_n\|_{G\to \Bbb R}=1$$ (to do this, one needs to apply Hahn-Banach where the dominating convex functional is the lower convex envelope of $$x\mapsto \min\{\|x_n\|_E^{-1} \|x\|_E, \|x\|_G\}$$).

Then we claim that $$\sup_n f_n = f$$ on $$D(A)$$. Indeed, on any separable Banach space it is true that if $$\{x_n\}$$ is a dense subset of the unit sphere and $$f_n$$ are linear functionals on $$B$$ such that $$f_n(x_n) = 1$$ and $$\|f_n\|=1$$ then $$\sup_n f_n = \|\cdot\|$$. See my answer here for the proof.

• I don't see why proving the measurability of $f$ is sufficient. On its face it seems that it only proves that the pre-images of open balls centered at the origin are measurable, and those do not generate the Borel $\sigma$-algebra. – Nate Eldredge May 3 at 20:11
• I have in mind the example in $\mathbb{R}$ of $g = 1_N - 1_{N^c}$ where $N$ is non-measurable. Then $|g|$ is measurable but $g$ is not. – Nate Eldredge May 3 at 20:12
• I made a clarification that both $E$ and $F$ should be separable. – lye012 May 3 at 20:56
• @NateEldredge no but the point here is that $A$ is linear so you can exploit that. See the modification above (and thanks for pointing out that detail). – Shalop May 3 at 21:54
• Sorry there was a bunch of notation in this answer that made no sense before, but it should be fixed now. – Shalop May 3 at 22:55