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?

  • $\begingroup$ What is the distinction between $l_{\infty}$ and $\mathbb{R}^{\infty}$? By $\ell^{\infty}$ I just mean the set of all sequence of real numbers indexed by $\mathbb{N}$, which I'm guessing is what is meant by $\mathbb{R}^{\infty}$. This problem is from a very old set of notes (1970s) so maybe the notation is non-standard. I'm a newcomer to measure theory so I could be wrong. $\endgroup$ – Akhil Jalan Nov 30 '18 at 20:37
  • $\begingroup$ $\ell^\infty$ usually stands for the set of all bounded real sequences. $\mathbb{R}^\infty$ is set of all real sequences, it is also denoted by $\mathbb{R}^{\mathbb{N}}$. $\endgroup$ – mechanodroid Nov 30 '18 at 21:20

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.

  • $\begingroup$ Clarifying questions: 1. Did you mean to say $\|x-y\| \geq 1$ for distinct $x, y \in X$? 2. When you say $X$ is Lebesgue, do you mean Lebesgue-measurable? And, to double-check, this is because all open/closed sets are part of the Borel $\sigma$-algebra? $\endgroup$ – Akhil Jalan Dec 1 '18 at 0:53
  • 1
    $\begingroup$ @AkhilJalan $x$ and $y$ are binary sequences and $||\cdot||$ is the supremum norm. If they are distinct, they differ at some index where one of them is one and the other is zero, so $||x-y||=1$. $\endgroup$ – Guacho Perez Dec 1 '18 at 0:54
  • $\begingroup$ Gotcha, thanks. And I assume the metric generated by the supremum norm is what defines open/closed sets here? $\endgroup$ – Akhil Jalan Dec 1 '18 at 0:58
  • 1
    $\begingroup$ @AkhilJalan yes, and these in turn generate the Borel and Lebesgue sets. $\endgroup$ – Guacho Perez Dec 1 '18 at 1:00

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