Let $\{I_{n}\}_{n \in \mathbb{N}}$ be the sequence of real intervals $[0, 1/2], [1/2, 1], [0, 1/3], [1/3, 2/3]$, and so on. Let $f_{n}: \mathbb{R} \rightarrow \mathbb{R}$ be the indicator function on $I_{n}$. Let $f:\mathbb{R} \rightarrow \ell^{\infty}$ be $f(x) = (f_{1}(x), f_{2}(x), ...)$.
Show that $f$ is not Lebesgue measurable.
I'm not sure where to start with this. I tried a contradiction proof, supposing that there exists a sequence of simple measurable functions which converge pointwise to $f$ (which would imply that $f$ is measurable). However, I'm having trouble reaching a contradiction.
My intuition for this approach is that the most "natural" such sequence, namely $g_{n}: \mathbb{R} \rightarrow \ell^{\infty}$ where $g_{n}(x) = (f_{1}(x), f_{2}(x), ..., f_{n}(x), 0, 0, ...)$ does not work, because for any $x \in [0, 1]$ there are infinitely many intervals $I_{k}$ such that $x \in I_{k}$, so for any $g_{n}$, we can show $\| g_{n}(x) - f(x) \| = \frac{1}{m} + \frac{1}{m+1} + ... = \infty$ for some $m \in \mathbb{N}$.
However, it is not enough to show that the particular sequence of simple measurable functions $(g_{n})$ fails to converge pointwise to $f$.
Is there some way to fix this approach? Or should I try something else entirely?
Let $X\subset \{0,1\}^{\Bbb N}$ and notice $X$ cannot have limit points because a limit point requires a Cauchy sequene within $X$, but $||x-y||=1$ for distinct $x,y \in X$. Hence, $X$ is vacuously closed, thus Lebesgue.
This means that $A=f(V)\subset \{0,1\}^{\Bbb N}$ is measurable, for $V\subset [0,1]$. Clearly, $V\subset f^{-1}(f(V))$, suppose $y \in f^{-1}(f(V))$ so that $f(y)=f(x)$ for $x\in V$. This implies $f_n(y)=f_n(x)$, which happens only if $y \in I_n$ for all $I_n \in J = \{I \in \{I_n\} : x\in I\}$, i.e. $y \in \bigcap_{I\in J}I$. But $\text{diam}(I_n)\to 0$ as $n\to \infty$, so $d(x,y)\to 0$ shows $y=x\in V$ and thus $f^{-1}(A)=V$. Set $V$ equal to the Vitali set to get the result.