Let $Y,X_1,X_2,\dots ,X_k$ be random vectors defined on a probability space $(\Omega,\mathcal F,P)$ and taking values in $\mathbb R^n$. Given $\omega\in\Omega$, let $\mathcal M(\omega)=\text{span}\{X_1(\omega),\dots,X_k(\omega)\}$, and define $\hat{X}(w)$ by
$$\|Y(\omega)-\hat{X}(w)\|=\inf_{v\in M(\omega)} \|Y(\omega)-v\|$$
Note that $\hat{X}(w)$ is well-defined by the projection theorem. Is the map $\omega\mapsto \hat{X}(w)$ $\mathcal F$-measurable?
EDIT: Using Tomas comment we can note that
$$\hat{X}=X[X'X]^{+}X'Y, $$
where $X:=[X_1,X_2,\dots ,X_k]$ is a random $n\times k$ matrix and $[X'X]^{+}$ denotes the Moore-Penrose inverse of $X'X$. From here we have that $A^{+}$ is a measurable function on the space of $k\times k$ matrices $A$, and therefore we get that each component of $[X'X]^{+}$ is $\mathcal F$-measurable. Then it follows that $\hat{X}$ is $\mathcal F$-measurable, since measurability is preserved under matrix multiplication.
Is this argument correct? Thanks a lot for your help.
