Is a subspace of functions that essentially depend only on one variable closed?

Question

Let $S$ be the subspace

$$\left\{f\in L^p( I^2)|\exists g\in L^p( I), f(x,y)=g(x), \mbox{a.e. } (x,y)\in I^2\right\}.$$

Is $S$ closed under the $L^p$ norm?

If I find some $h$ in the dual $L^q$ ($p+q=1$) to separate a point and a convex set in this subspace $S$, could I show that $h$ is also essentially depend only on one variable?

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I think the first step would be to suppose there exists $\{f_n\}_n \subset S$ such that $f_n\to L^p\phantom{0} f\in L^p(I^2)$. Then there exists $\{g_n\}_n \subset L^p(I^2)$ such that $g_n\to L^p \phantom{0} f\in L^p(I^2)$? Is it right?

Then, we have for any $\epsilon>0$ and any $n\geq N$, $||g_n-f||_p\leq \epsilon$. Now suppose $f\notin S$. There is positive measure $E$ in $I^2$ that $f=h(x,y)$ on $E$? We can choose $f(x,y)=g_n(x)$ on $I^2\setminus E$, then $||g_n-f||_p>\epsilon$ and we have contradiction?

Answer 1

Suppose that $(f_n)_{n\geqslant 1}$ is a sequence of elements of $S$ converging in $\mathbb L^p(I^2)$ to some function $f$. Then we extract a subsequence $(f_{n_k})_{n\geqslant 1}$ which converges almost everywhere to $f$. There exists a function $g_{n_k}$ such that for almost every $(x,y)\in I$, $f_{n_k}(x,y)=g(x)$. Thus, for almost every $(x,y)\in I$, we have $$f(x,y)=\lim_{k\to \infty}f_{n_k}(x,y)=\lim_{k\to \infty}g_{n_k}(x).$$ Defining $g(x):=\lim_{k\to \infty}g_{n_k}(x)$ and $g(x)=0$ outside the null set where the convergence of $(f_{n_k})$ does not take place, we derive that $f\in S$.

Answer 2

For a quick proof, observe that the map $T: L^p(I) \to L^p(I^2)$ that sends $g$ to such an $f$ is a linear map that preserves norms, so it can be thought of as an isometric embedding. An isometrically embedded image of a complete space is always closed in the ambient space.