*This is a follow-up to [this previous question](https://mathoverflow.net/q/434802/102946), but under stronger assumptions.*


Let $(X, \mu)$ be a (say, $\sigma$-finite) measure space, let $g \in L^2$ (say, over the real
scalar field). Let $\tilde g: X \to \mathbb{R}$ be a measurable function and assume that there exists a norm dense vector subspace $D$ of $L^2$ with the following two properties:


1. The subspace $D$ is an *operator range* in $L^2$, i.e., there exists a complete norm on $D$ under which the injection $D \hookrightarrow L^2$ is continuous.
2. For every $f \in D$ the function $\tilde g f$ is integrable, and we have $\int \tilde g f \, d\mu = \int g f \, d\mu$.


**Question.** Does it follow that $\tilde g = g$ almost everywhere?


**Remarks.**


* Without the assumption that $D$ be an operator range, the answer is **no**. Two counterexamples were given by Piotr Hajlasz and Gro-Tsen in the answers to question linked at the beginning of the post.
* The complete norm on $D$ is not required to render $D$ a Hilbert space (although this might be an interesting variant of the question).
* As explained in the previous question, it suffices to show that $\tilde g \in L^2$.
* As also explained in the previous question, the answer is **yes** if $D$ is a *lattice ideal* in $L^2$ (even if $D$ is not assumed to be an operator range).



Representing an $L^2$-functional by a non-$L^2$-function on a dense subspace - Part II
Gro-Tsen's answer to your previous question provides a counterexample if you define $D$ to be all vectors in $\ell\_2$ that are of the form $\sum\_n a\_n f\_n$, where
$f\_n = e\_n + e\_{n+1}$, $(e\_n)$ is the unit vector basis for $\ell\_2$, and $\sum |a\_n| < \infty.$ $D$ is the range of a bounded linear operator from $\ell\_1 $ into $\ell\_2$.


