# Show that $f: \mathbb{R}: \rightarrow \ell^{\infty}$ is not Lebesgue integrable?

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?

• 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. – Akhil Jalan Nov 30 '18 at 20:37
• $\ell^\infty$ usually stands for the set of all bounded real sequences. $\mathbb{R}^\infty$ is set of all real sequences, it is also denoted by $\mathbb{R}^{\mathbb{N}}$. – 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.
• 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? – Akhil Jalan Dec 1 '18 at 0:53
• @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 $||x-y||=1$. – Guacho Perez Dec 1 '18 at 0:54