Let $L$ be an $L^p$-space, $1<p<\infty$, associated with an arbitrary measure space $(X, \mathcal F, \mu)$. A semi-partition of $X$ is a finite disjoint collection of measurable sets with finite measures. Let $A$ be the set of all semi-partitions of $X$. We introduce a partial order into $A$ as $\alpha\leq \alpha'$ meaning that each set in $\alpha$ is a union of some sets in $\alpha'$, this makes $A$ a directed set. Let $E_\alpha:L\to L$ be the conditional expectation operator with respect to the semi-partition $\alpha$, mapping functions to their average values on the sets of $\alpha$ and to zero outside of these sets. We have that for fixed $f\in L$, the net $\{E_\alpha f\}$ converges to $f$. Let $L_\alpha$ be the range of $E_\alpha$ (this is a finite dimensional $L^p$-space). Let $\mathcal U$ be an ultrafilter on $A$ containing all sets $\{\alpha\in A: \alpha\geq\alpha_0\}$ for arbitrary $\alpha_0\in A$.
I would like to know if the ultraproduct $W$ of the $L_\alpha$ with respect to $\mathcal U$ is isometric isomorphic to $L$ itself. If $A$ contains a largest element this is true of course, but my intuition says it should be also true otherwise.
Here is what I have tried:
Let $\phi: L\to W, f\mapsto (E_\alpha f)_\mathcal U$, where $(x_\alpha)_\mathcal U$ denotes the equivalence class of $(x_\alpha)_{\alpha\in A}$ with respect to $\mathcal U$.
The mapping is isometric: $\|(E_\alpha f)_\mathcal U\| = \lim_\mathcal U\|E_\alpha f\|$. Since $\|E_\alpha f\|$ net-converges to $\|f\|$ this is also true for the ultrafilter limit due to the property we demanded of our ultrafilter.
I have problems with the surjectivity. Let $(f_\alpha)_\mathcal U\in W$ be arbitrary. My idea was to define a functional on $\varphi$ on $L^q(X)$ via $\varphi(g) = \lim_\mathcal U\int_{L_\alpha}(E_\alpha g)f_\alpha$. The limit exists because $\int_{L_\alpha}(E_\alpha g)f_\alpha$ is bounded. This also shows that $\varphi$ is bounded, hence there is $f\in L$ which induces $\varphi$. I think that we should have $(E_\alpha f)_\mathcal U = (f_\alpha)_\mathcal U$, but I can't show that $\lim_\mathcal U\|E_\alpha f-f_\alpha\| = 0$. If $(f_\alpha)_\mathcal U$ is already of the form $(E_\alpha f)_\mathcal U$ then this construction with the functional indeed gives back $f$, but in the general case I don't see it. Can anybody help me?
I think that the answer is NO since an ultraproduct of $L^p$-spaces $L^p(\Omega_i)$ is an $L^p$-space on a measure space $\Omega^{\mathcal{U}}$ and such a measure space is a "bigger" measure space and generally no $\sigma$-finite even if the $\Omega_i$'s are $\sigma$-finite.