Riesz map on $L^2(0,T;H)$ -- it's not "unique" in a way

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Let $R:L^2(0,T;H) \to L^2(0,T;H^*)$ denote the Riesz representation map.

Given $u \in L^2(0,T;H)$, $Ru \in L^2(0,T;H^*)$ can be changed in $[0,T]$ on a set of measure zero. So $Ru$ is not unique in that sense. Of course it is unique in the sense of $L^p$ spaces since they consist of equivalence classes.

But at the end of the day, when I put in $u$, $Ru$ should give me a function, not an equivalence class. Which function out of those should I take?

Suppose I always want to take the function $Ru$ such that $(Ru)(0)= 0$. Is it OK?

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Even $u\in L^2([0,T];H)$ is an equivalence class (not a function), so is $Ru$. However sometimes there exists a "good" candidate (i.e. a function in the equivalence class $Ru$ with some "more desirable" properties, such as regularity...). Imposing $Ru(0)=0$ is very arbitrary, and there is no general criteria to decide which function is preferable in the equivalence class (e.g. in the class of functions $\{f:f(x)=1\ for\ a.e.\ x\}$ the "good" candidate is $f\equiv 1$, so $f(0)=1$).