I'm considering the following problem. Let $s\in L^\infty ((0,T)\times K)$ for some compact $K\subset \mathbb{R^n}$ be given. Consider the Steklov average in time of $s$, i.e. for $h>0$ and $t\in(0,T-h)$ define
$$s_h(x,t)=\int_t^{t+h} s(x,\tau) d\tau$$ and choose $s=0$ for $t>T$. Now let $f$ be some smooth function in $(0,T)\times K$. Due to the Steklov average $s_h \in H^1(0,T;L^\infty(\Omega))$. Hence it is allowed to write for almost every $sigma\in (0,T-h)$
$$\int_K s_h(x,\sigma)f(x,\sigma) dx $$.Of course I want to send $h\to 0$. Now $s_h\to s$ at least in $L^2((0,T)\times \Omega)$. So for a subsequence we find pointwise convergence. Then formally we find for that subsequence and a.e. $t\in (0,T)$ that $$\int_K s_h(x,\sigma)f(x,\sigma) dx \to \int_K s(x,t) f(x,t) $$.
But I don't get the justification for this: $s$ does not have a trace so it can be changed arbirtrarily in $t$. So the integral on the right hand side might be different for any representative of $s$ might even depend on the subsequence. (for each representative the subsequence might be different I think).
Any clues on that or do I understand something wrong here?
Almost every point of $(0,T)\times K$ is a Lebesgue point for $s$. By Fubini's theorem, for almost every $t$ the set of Lebesgue points has full measure in $\{t\}\times K$. This allows the trace of $s$ on $\{t\}\times K$ to be defined in a natural way, using some form of limit of averages as in your post.
The set of such $t$ depends on function $s$, of course. If you fix $t$ and vary $s$ over $L^\infty((0,T)\times K)$, there is no natural way to assign an element of $L^\infty(\{t\}\times K)$ to every element of $L^\infty((0,T)\times K)$.