$L^2(\mu)$ equivalent scalarproduct Let $\langle \cdot, \cdot \rangle_{L^2(\mu)}$ denote the usual scalar product on $L^2(\mu)$. Let $\langle \cdot, \cdot \rangle_{\varphi}$ denote a different scalar product on $L^2(\mu)$ wich is norm equivalent to $\langle \cdot, \cdot \rangle_{L^2(\mu)}$. 
Is the scalar product $\langle \cdot, \cdot \rangle_{\varphi}$ induced by a measure? 
If $  \lVert \cdot \rVert _{\varphi}$ is just a Norm this does not hold. There is a counter example if $\mu$ is finite meassure, $L^2(\mu) \subset L^1(\mu)$ holds. Therefor you can simply add the $L^1(\mu)$ Norm to the usual $L^2(\mu)$ Norm and this cannot be induced by a meassure.
 A: Typically, $\langle \cdot, \cdot \rangle_\varphi$ need not be induced by a measure.  For instance, an inner product induced by a measure has the property that functions of disjoint support are orthogonal, and this property need not be preserved for equivalent inner products.
For an explicit counterexample, let $X = \{a,b\}$ be a space with two points and let $\mu$ be counting measure, so that $L^2(X,\mu)$ is isomorphic in the obvious way to $\mathbb{R}^2$ with its usual dot product, and the usual orthonormal basis $\{(1,0), (0,1)\}$ for $\mathbb{R}^2$ corresponds to the basis $\{1_{\{a\}}, 1_{\{b\}}\}$ for $L^2(\mu)$. Consider the invertible matrix $T = \left(\begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix}\right)$ and the new inner product $\langle v,w \rangle_{\varphi} := \langle Tv, Tw \rangle$.  This is easily checked to be equivalent to the original inner product; indeed, all inner products on $\mathbb{R}^2$ are equivalent because it is finite dimensional.
Then $\langle 1_{\{a\}}, 1_{\{b\}} \rangle_\varphi = \langle 1_{\{a\}}, 1_{\{a\}} + 1_{\{b\}} \rangle = 1$.  But for any measure $\nu$, we have $\int_X 1_{\{a\}} 1_{\{b\}} \,d\nu = \int_X 0 \,d\nu = 0$.
You can create a similar example on any measure space $(X,\mu)$ which is not too trivial, by letting $f,g$ be any two functions of disjoint support, neither of which is $\mu$-a.e. zero.  Let $Tv = v + \langle v,g \rangle f$ and set $\langle v,w \rangle_{\varphi} := \langle Tv, Tw \rangle$.  This again will be an inner product equivalent to the original one (because $T$ is invertible) and has $\langle f, g \rangle_{\varphi} \ne 0$.
