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A basic result in real analysis is that any measurable function is an a.e. limit of a step function sequence (yet a pointwise limit of a simple function sequence), but the statement does not hold when the “a.e.” is replaced with everywhere. My question is how to find a counterexample to the “everywhere” statement.

My attempt:
I’ve tried to use the fact that a step function is different from a simple one in that it is continuous on the complement of a zero-measure set, then maybe apply the Egorov’s thm. Considering this we are motivated to choose an everywhere discontinuous characteristic function of some “bad” measurable set. But then I got stuck, since once a.e. is involved, it seems hard to dispense with it (so as to arrive at an counter argument).

Please help, thx!

Juggler
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  • What's your definition of a step function? – Clement Yung Nov 20 '21 at 03:14
  • Do you consider points/lines etc (e.g. singletons and ${0} \times [0,1]$) closed cubes? – Clement Yung Nov 20 '21 at 03:25
  • @ClementYung Sorry I was a bit careless. It’s a linear combination of characteristic functions of closed rectangles, and we allow the rectangles to degenerate (i.e. with empty interior). – Juggler Nov 20 '21 at 03:34
  • I am not sure if this is what you want. However a simple example is a function $=1$ for irrational points and $=0$ for rational points. – herb steinberg Nov 20 '21 at 04:34
  • @herbsteinberg Thank you, but I think Dirichlet function can be a limit of step functions. You just list rational points and construct disjoint small intervals to separate some of them at every finite stage. – Juggler Nov 20 '21 at 04:59
  • @Juggler I don't see the everywhere convergence. – herb steinberg Nov 20 '21 at 05:22
  • @herbsteinberg You are right. I’m sorry. But we can just add a term corresponding to a single rational point every time and it follows that Dirichlet’s function is a limit of step functions. So is the difference between it and constant 1. – Juggler Nov 20 '21 at 05:36
  • @Juggler The rational points (in this construction) are one point step functions? – herb steinberg Nov 20 '21 at 18:50
  • @herbsteinberg Yes, and every finite stage we only have a finite and hence discrete subset. – Juggler Nov 20 '21 at 22:40

2 Answers2

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I believe I have an answer. Note that step functions are all Borel-measurable, and a limit of Borel-measurable is also Borel-measurable. So take the characteristic function of a Lebesgue-measurable but non-Borel-measurable set and we have an example.

Juggler
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It is possible to prove that there exists a Borel-measurable function which is not the everywhere-limit of any step function.

If $(f_n)$ is a sequence of real-valued functions on a set $X$, then its point of convergence is given by

$$E=\bigcap_{\varepsilon\in\mathbb{Q}_{>0}}\bigcup_{N\geq1}\bigcap_{m,n\geq N}\{x\in X:|f_m(x)-f_n(x)|<\varepsilon\}.$$

(Note that $E$ is precisely the set of all $x$ at which $(f_n(x))_{n\geq1}$ is a Cauchy sequence in $\mathbb{R}$.)

Now, if $(f_n)$ is any sequence of step functions, then $\{x\in X:|f_m(x)-f_n(x)|<\varepsilon\}$ is a finite union of intervals. So, it is a $G_{\delta}$-set and hence belongs to the class $\mathbf{\Pi}_{2}^{0}$ in the Borel hierarchy on $\mathbb{R}$.

From this, we know that $E$ is an $G_{\delta\sigma\delta}$-set, and so, it lies in the class $\mathbf{\Pi}_{4}^{0}$. Since we know that there exists a Borel set $B$ which is not in the class $\mathbf{\Pi}_{4}^{0}$, the indicator function $\mathbf{1}_{B}$ is Borel-measurable but cannot be an everywhere-limit of any sequence of step functions.

Although I have little expertise in the descriptive set theory, it seems that no "natural" examples of Borel sets outside of $\mathbf{\Pi}_4^{0}$ is known in the literature. (See the comment in Section 23 of Kechris.) So, such set has to be artificially engineered. One such argument can be found in this post.

Sangchul Lee
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