Completeness of variation on $L_2$-space I'm trying to show completeness of the space of measurable functions $f:[0,1]\rightarrow \mathbb C$ with the condition $\int|f|^2kd\mu<\infty$.  $k$ is a measurable function, $k:[0,1]\rightarrow[0,\infty)$, that vanishes on a negligible set. we define inner product on this space by $(f,g)=\int f\bar{g}kd\mu$. I tried variations on the standart proof of completeness in $L_2$, but it failed. Any ideas? 
 A: Define $\nu (E)=\int_E k d\mu$. Then $\nu$ is a measure and $L^{2} (\nu)$ is complete. This is exactly the space you are considering. 
A: Let $\mathcal H$ be your inner-product space, with your inner product, and let $L_2([0,1])$ denote the standard $L_2$ Hilbert space on $[0,1]$, with the standard inner product. Why not show that $\mathcal H$ and $L_2[0,1]$ are isomorphic as inner product spaces? Because if two inner products spaces are isomorphic, and one of them is complete, then so is the other one.
Indeed, the map $\psi : \mathcal H \to L_2([0,1])$ sending $ f \mapsto f \sqrt{k} $
is an isomorphism of Hilbert spaces, and it should be possible to verify this. When doing so, be aware of the following:


*

*Elements in $\mathcal H$ in $L_2([0,1])$ are only defined up to equality almost everywhere, so you need to check if two functions $f_1, f_2$ agree almost everywhere, then $\psi(f_1), \psi(f_2)$ also agree almost everywhere. This is necessary in order for $\psi$ to be well-defined.

*To see that $\psi$ is surjective, we must make use of the fact that $k$ vanishes on a null set. Indeed, pick a $g \in L_2([0,1])$: can we find an $f \in \mathcal H$ such that $\psi(f) = g$? An obvious choice would be to take $f = g / \sqrt{k}$. But $g / \sqrt{k}$ isn't actually defined at those points where $k $ vanishes. Fortunately, the set of points where $k$ vanishes is null, and functions in $\mathcal H$ only need to be defined almost everywhere, so we can get away with this!
