$L^2$ representation of Hilbert subspace of $L^2$

32 Views Asked by At

Suppose that $\mathcal{H}$ is a Hilbert space which is a subspace of $L^2(X,\mathcal{A},\mu)$. Does there exist a sub-$\sigma$-algebra $\mathcal{B}$ of $\mathcal{A}$ such that $\mathcal{H}=L^2(X,\mathcal{B},\mu |_\mathcal{B})$?

1

There are 1 best solutions below

1
On BEST ANSWER

Let $f$ be a bounded function in $L^{2}(X,\mathcal A,\mu)$ such that $f^{2}$ is not a scalar multiple of $f$. Let $M$ be the one dimensional subspace generated by $f$. If $M$ has the desired form then $f^{2}$ is an $L^{2}$ function measurable w.r.t. $\mathcal B$ so it must belong to $M$. This is a contradiction.