[proofplan]
We prove the operator identity by testing both sides on an arbitrary element of $L^2(\mathbb{R})$. Since the Weyl operators are bounded operators defined by explicit pointwise formulas on representatives, it suffices to compare the two resulting $L^2$-classes by evaluating their representatives for $\mathcal{L}^1$-a.e. real $x$. The only computation is the phase shift caused by translating before multiplying versus multiplying before translating.
[/proofplan]
[step:Evaluate $U(a)V(b)$ on an arbitrary $L^2$ function]
Fix $a,b \in \mathbb{R}$. Let $\psi \in L^2(\mathbb{R})$, and choose a measurable representative, still denoted $\psi: \mathbb{R} \to \mathbb{C}$, of its $\mathcal{L}^1$-a.e. equivalence class. By the definition of $V(b)$ and then the definition of $U(a)$, for $\mathcal{L}^1$-a.e. $x \in \mathbb{R}$,
\begin{align*}
[U(a)V(b)\psi](x) = e^{iax}[V(b)\psi](x)
\end{align*}
and hence
\begin{align*}
[U(a)V(b)\psi](x) = e^{iax}\psi(x-b).
\end{align*}
[/step]
[step:Evaluate $e^{iab}V(b)U(a)$ on the same function]
Using first the definition of $U(a)$ and then the definition of $V(b)$, for $\mathcal{L}^1$-a.e. $x \in \mathbb{R}$,
\begin{align*}
[V(b)U(a)\psi](x) = [U(a)\psi](x-b).
\end{align*}
Therefore
\begin{align*}
[V(b)U(a)\psi](x) = e^{ia(x-b)}\psi(x-b).
\end{align*}
Multiplying by the scalar $e^{iab} \in \mathbb{C}$ gives
\begin{align*}
[e^{iab}V(b)U(a)\psi](x) = e^{iab}e^{ia(x-b)}\psi(x-b).
\end{align*}
Since $iab + ia(x-b) = iax$, this becomes
\begin{align*}
[e^{iab}V(b)U(a)\psi](x) = e^{iax}\psi(x-b).
\end{align*}
[guided]
The only point at which the commutation relation appears is the comparison of phases. We first apply $U(a)$ to $\psi$, which means multiplication by the phase function $x \mapsto e^{iax}$. Thus the representative of $U(a)\psi$ is the measurable function
\begin{align*}
x \mapsto e^{iax}\psi(x).
\end{align*}
Now applying $V(b)$ translates the input by $b$, so for $\mathcal{L}^1$-a.e. $x \in \mathbb{R}$,
\begin{align*}
[V(b)U(a)\psi](x) = [U(a)\psi](x-b).
\end{align*}
Substituting the formula for $U(a)\psi$ at the point $x-b$ gives
\begin{align*}
[V(b)U(a)\psi](x) = e^{ia(x-b)}\psi(x-b).
\end{align*}
The Weyl relation says that this is not exactly the same as $U(a)V(b)\psi$, but differs by the constant phase factor $e^{iab}$. Multiplying the last display by $e^{iab}$ yields
\begin{align*}
[e^{iab}V(b)U(a)\psi](x) = e^{iab}e^{ia(x-b)}\psi(x-b).
\end{align*}
The exponent simplifies by ordinary algebra:
\begin{align*}
iab + ia(x-b) = iab + iax - iab = iax.
\end{align*}
Therefore, for $\mathcal{L}^1$-a.e. $x \in \mathbb{R}$,
\begin{align*}
[e^{iab}V(b)U(a)\psi](x) = e^{iax}\psi(x-b).
\end{align*}
This is exactly the representative obtained from $U(a)V(b)\psi$.
[/guided]
[/step]
[step:Conclude equality of the bounded operators]
The two computations give, for the arbitrary $\psi \in L^2(\mathbb{R})$ and for $\mathcal{L}^1$-a.e. $x \in \mathbb{R}$,
\begin{align*}
[U(a)V(b)\psi](x) = [e^{iab}V(b)U(a)\psi](x).
\end{align*}
Thus $U(a)V(b)\psi$ and $e^{iab}V(b)U(a)\psi$ represent the same element of $L^2(\mathbb{R})$. Since $\psi \in L^2(\mathbb{R})$ was arbitrary, the two operators agree on all of $L^2(\mathbb{R})$:
\begin{align*}
U(a)V(b) = e^{iab}V(b)U(a).
\end{align*}
This proves the Weyl form of the canonical commutation relations for the Schrödinger Weyl operators.
[/step]
