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 $\pi_F$.

  • $\begingroup$ @Nate Eldredge I see, thanks a lot for the help! $\endgroup$ – 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.

  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge May 3 at 20:11
  • $\begingroup$ 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. $\endgroup$ – Nate Eldredge May 3 at 20:12
  • $\begingroup$ I made a clarification that both $E$ and $F$ should be separable. $\endgroup$ – lye012 May 3 at 20:56
  • $\begingroup$ @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). $\endgroup$ – Shalop May 3 at 21:54
  • $\begingroup$ Sorry there was a bunch of notation in this answer that made no sense before, but it should be fixed now. $\endgroup$ – Shalop May 3 at 22:55

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