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Let $(\Omega, \mathcal F) $ be a measurable space; $A:\Omega \to \mathbb R^{n\times n} $ be a measurable matrix-valued function and $b:\Omega\to \mathbb R^n$ be a measurable vector-valued function. Assume that $A(\omega) $ is symmetric positive semi-definite and $b(\omega) \in \operatorname{Range}(A(\omega)) $ for all $\omega\in\Omega$. Hence for each $\omega$ there exists a $x(\omega)$ such that $A(\omega) x(\omega) =b(\omega) $. However, I'm wondering about:

Problem. Does there exist a choice of $x:\Omega\to\mathbb R^n$ that is measurable?

Attempts. I've thought of using Moore-Penrose inverse $A^+$, but I'm not sure how it gives a measurable function $x$. Since $x=A^+b$ is a solution and the minimizer of the corresponding linear least square problem, maybe this boils down to finding a measurable selection, for example this result comes to mind. However, I'm not sure how I should apply it, I couldn't show the requirements.

Another way is to get $x$ via Gauss-elimination. The operations involved seem to yield a measurable $x$, but at some point there might be infinitely many choices, so I'll be stuck at the same position of having to choose a measurable one among those solutions.

I also thought of finding the eigenvalues decomposition of $A$. Namely writing $A=Q^\top \Lambda Q$, then it's not difficult to solve for $x$ and it yields measurability as long as the decomposition of $A$ is measurable, i.e. $Q$ and $\Lambda$ are both measurable. I've heard of algorithms but there were for non-singular matrices with non-repeating eigenvalues and many restrictions. This question is related, but the argument seems non-trivial to me, I believe there is an easier way.

Finally, I am sure a measurable choice exists, because I have seen this in the construction of a Brownian motion when given a quadratic variation absolutely continuous with respect to the Lebesgue measure. However, all the references I'm aware of say something along "those mappings are obviously measurable".

Any help, reference or hint is appreciated. Thanks in advance.

Shashi
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  • Maybe I'm missing something obvious (which is why I'm leaving this as a comment), but why doesn't $\omega \mapsto A_\omega^+ b_\omega$ work? After all, this function is measurable (if you like, each entry of the output vector is a polynomial in the entries of $A$ and $b$) and solves your problem (since we're assuming $A_\omega x=b_\omega$ has a solution for each $\omega$)... I'm not sure why a selector is necessary. – HallaSurvivor Apr 25 '21 at 05:40
  • @HallaSurvivor I didn't know how to get the Moore-Penrose inverse. I do know that the real inverse satisfies what you said, but that may not exist. Do you have a reference of what you stated? – Shashi Apr 25 '21 at 11:04
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    @Shashi Maybe we can use the limit relation of the pseudoinverse (which is denoted as Tikhonov regularization in your link) since the limit of the measurable function is also measurable. – Needmoremath Apr 25 '21 at 13:10
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    @night_owl wow indeed, I've missed that!! Thank you very much! – Shashi Apr 25 '21 at 13:25
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    There is a saying: "If you can write it down, it's measurable." (Obviously, not "true," but helpful to keep in mind.) –  Apr 25 '21 at 14:13
  • I'm glad I helped! @PeterMorfe Exactly, that quote is my belief :> – Needmoremath Apr 25 '21 at 18:54

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