Show measurability of parameterized function I'm working on exercise 1 of chapter 4 of Brian C. Hall's Quantum Theory for Mathematicians. I'm most of the way through it, but at the very end, I need to show that a certain complex valued function on $\mathbb{R}^2$ is locally integrable. I believe that the first step then is to prove the function is measurable.
We are given a function $\psi_0\in L^2(\mathbb{R})$. Let $\mathscr{F}$ represent the Fourier-Plancherel transform $\mathscr{F}:L^2(\mathbb{R})\to L^2(\mathbb{R})$. For any function $f\in L^2(\mathbb{R})$, we write $\hat{f}$ for $\mathscr{F}(f)$. For a parameter $t\in\mathbb{R}$, define
\begin{equation}\tag{1}
  \theta_t(k)=\hat{\psi}_0(k)e^{-i\hbar k^2t/(2m)}\qquad(k\in\mathbb{R}).
\end{equation}
We see that $\theta_t\in L^2(\mathbb{R})$, so we then define
$$\psi_t=\mathscr{F}^{-1}(\theta_t)\qquad (t\in\mathbb{R}).$$
Finally, define $\psi:\mathbb{R}^2\to\mathbb{C}$ by
$$\psi(x,t)=\psi_t(x).$$
It is required to show that $\psi$ is locally integrable, so I want to first show that it is measurable. It is clear that $\psi_t$ is measurable (on $\mathbb{R}$) for each $t\in\mathbb{R}$, but I haven't been able to demonstrate the $\mathbb{R}^2$-measurability of $\psi$.
So, what have I tried?
Let $U$ be an open subset of $\mathbb{C}$. Then $\psi^{-1}(U)=\bigcup_{t\in\mathbb{R}}(\psi_t^{-1}(U)\times\{t\})$, which, unfortunately, is an uncountable union of sets which are $\mathbb{R}^2$-measurable.
I don't see where to go from here.
Now, there is one other possibility. To go through that, I need to quote the first part of the problem statement exactly:

A locally integrable function $\psi(x,t)$ satisfies the free Schrödinger equation in the weak (or distributional) sense if for each smooth compactly supported function $\chi$, we have
  \begin{equation}\tag{4.30}
\int_{\mathbb{R}^2}\psi(x,t)\biggl[\frac{\partial\chi}{\partial t}+\frac{i\hbar}{2m}\frac{\partial^2\chi}{\partial x^2}\biggr]\,dx\,dt=0.
\end{equation}

The free Schrödinger equation is given earlier in the text as
\begin{equation}\tag{4.1}
\frac{\partial\psi}{\partial t}=\frac{i\hbar}{2m}\frac{\partial^2\psi}{\partial x^2}.
\end{equation}
Now (4.30) is written in a weird way. It starts out as a single integral over all of $\mathbb{R}^2$ but ends up as if it were an iterated integral ($dx\,dt$). If, as stated, $\psi$ is locally integrable, then it is assumed to be $\mathbb{R}^2$-measurable and it wouldn't matter how you interpreted the integral since the function in square brackets is smooth and compactly supported, so Fubini applies. But it seems like it would be possible to restate the problem as "Suppose $\psi_t$ is a locally integrable function on $\mathbb{R}$ for each $t\in\mathbb{R}$ and suppose we write $\psi(x,t)=\psi_t(x)$. Then $\psi$ satisfies ...". Then we would replace the single integral over $\mathbb{R}^2$ with iterated integrals each over $\mathbb{R}$.
I haven't gone far enough in the text to know if this interpretation would run into problems (mathematically or physically), but it would certainly eliminate the problem of showing that $\psi$ was $\mathbb{R}^2$-measurable. If someone can help with a proof that $\psi$ is $\mathbb{R}^2$-measurable, then I won't have to go there. On the other hand, if such a proof is not possible, I don't see an alternative but to go there.
 A: Well, after much thrashing and help from two MIT professors, I finally have an answer. I concluded that if you modify the question to be slightly more reasonable, then the answer is affirmative.
So, the question as written above is OK through equation (1), and $\theta_t$ is indeed
in $L^2(\mathbb{R})$ for all $t\in\mathbb{R}$. But it is a misstep to define $\psi_t$
and $\psi$ as I originally wrote it. Here is the proper form of the problem:
Show there exists a measurable function $\Psi:\mathbb{R}^2\to\mathbb{C}$ such that for
each $t\in\mathbb{R}$:
(a) $\Psi_t\in L^2(\mathbb{R})$, where $\Psi_t(x)=\Psi(x,t)$ for all $x\in\mathbb{R}$.
(b) $\mathscr{F}(\Psi_t)=\theta_t$, and
(c) $\Psi_0=\psi_0$ as members of $L^2(\mathbb{R})$.
The proof is a bit cumbersome, so I will only give some highlights here, and I won't bother with the normalization constants in Fourier transforms.
By dominated convergence,
\begin{equation}\tag{2}
  \lim_{j\to\infty}\lvert\lvert\,\lvert\hat{\psi}_0\rvert^2\chi_{[-j,j]^c}\,\rvert\rvert_1=0.
\end{equation}
By (2), we can inductively define a strictly increasing sequence of integers $A_1<A_2<\cdots$, such that for each $n=1,2,\dots$,
\begin{equation}\tag{3}
  \int_{\lvert k\rvert\geq A_n}\lvert\hat{\psi}_0(k)\rvert^2\,dk<4^{-n}.
\end{equation}
For $n=1,2,\dots$, define
\begin{equation}\tag{4}
  \Psi_n(x,t)=\int_{\lvert k\rvert\leq A_n}\!\theta_t(k)e^{ikx}\,dk
  =\int_{\lvert k\rvert\leq A_n}\!\hat{\psi}_0(k)e^{i(kx-\hbar k^2t/(2m))}\,dk.
\end{equation}
$\Psi_n$ is continuous on $\mathbb{R}^2$ by dominated convergence. Let
\begin{equation}\tag{5}
  C=\Bigl\{(x,t)\in\mathbb{R}^2:\lim_{n\to\infty}\,\Psi_n(x,t)\text{ exists and is finite}\Bigr\}.
\end{equation}
$C$ is a Borel subset of $\mathbb{R}^2$. Define $\Psi:\mathbb{R}^2\to\mathbb{C}$ by
\begin{equation}\tag{6}
  \Psi(x,t)=\lim_{n\to\infty}\bigl(\Psi_n(x,t)\chi_C(x,t)\bigr).
\end{equation}
$\Psi$ is Borel measurable on $\mathbb{R}^2$. For $n=1,2,\dots$, $t,k\in\mathbb{R}$, define
\begin{equation*}
  \Omega_{n,t}(k)=\theta_t(k)\chi_{[-A_n,A_n]}(k).
\end{equation*}
Then $\Omega_{n,t}\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ and by (4)
\begin{equation*}
  \Psi_{n,t}(x)=\Psi_n(x,t)=\int_{-\infty}^\infty\!\Omega_{n,t}(k)e^{ikx}\,dk.
\end{equation*}
By Rudin Real & Complex Analysis Theorem 9.14, with $f:=\mathscr{F}^{-1}(\Omega_{n,t})$,
\begin{equation*}
  (\mathscr{F}^{-1}(\Omega_{n,t}))(x)=\int_{-\infty}^\infty\!\Omega_{n,t}(k)e^{ikx}\, dk\qquad\text{a.e.},
\end{equation*}
so that $\Psi_{n,t}(x)=(\mathscr{F}^{-1}(\Omega_{n,t}))(x)$ for almost every $x\in\mathbb{R}$. Hence $\Psi_{n,t}\in L^2(\mathbb{R})$ and as members of $L^2(\mathbb{R})$, $\Psi_{n,t}=\mathscr{F}^{-1}(\Omega_{n,t})$. Using this, the Plancherel theorem and (3), we get that for $n=1,2,\dots$,
\begin{equation*}
  \lvert\lvert\Psi_{n+1,t}-\Psi_{n,t}\rvert\rvert_2<2^{-n}.
\end{equation*}
Thus, without having to take a subsequence (see proof of Rudin R&CA Theorem 3.11),
for each $t\in\mathbb{R}$, $\{\Psi_{n,t}(x)\}$ converges to a finite limit for almost
all $x\in\mathbb{R}$. From (5) and (6), $\Psi_{n,t}(x)\to\Psi_t(x)$ for almost
all $x\in\mathbb{R}$.
Let $\xi_t=\mathscr{F}^{-1}(\theta_t)$ for each $t\in\mathbb{R}$. By (4)
\begin{equation*}
  \Psi_{n,t}(x)=\int_{-A_n}^{A_n}\!\hat{\xi}_t(k)e^{ikx}\,dk,
\end{equation*}
so by the Plancherel theorem,
\begin{equation}\tag{7}
  \lvert\lvert\Psi_{n,t}-\xi_t\rvert\rvert_2\to 0.
\end{equation}
Since
$\lvert\lvert(\Psi_{n+1,t}-\xi_t)-(\Psi_{n,t}-\xi_t)\rvert\rvert_2
  =\lvert\lvert\Psi_{n+1,t}-\Psi_{n,t}\rvert\rvert_2<2^{-n}$,
as above $\{\Psi_{n,t}(x)\}$ converges to $\xi_t(x)$ for almost all $x\in\mathbb{R}$.
Combining the two subsets of measure $0$, $\Psi_t(x)=\xi_t(x)$ for almost all
$x\in\mathbb{R}$, so $\Psi_t\in L^2(\mathbb{R})$ and $\Psi_t=\xi_t$ in $L^2(\mathbb{R})$. Hence $\mathscr{F}(\Psi_t)=\mathscr{F}(\xi_t)=\theta_t$.
That takes care of (a) and (b). Finally, in $L^2(\mathbb{R})$,
$$\Psi_0=\xi_0=\mathscr{F}^{-1}(\theta_0)=\mathscr{F}^{-1}(\hat{\psi}_0)=\psi_0,$$
which proves (c).
