Roland Manning

2022-06-26

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. How to find a counterexample to the “everywhere” statement?

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).

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).

trajeronls

Beginner2022-06-27Added 21 answers

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

$E=\bigcap _{\epsilon \in {\mathbb{Q}}_{>0}}\bigcup _{N\ge 1}\bigcap _{m,n\ge N}\{x\in X:|{f}_{m}(x)-{f}_{n}(x)|<\epsilon \}.$

(Note that $E$ is precisely the set of all $x$ at which $({f}_{n}(x){)}_{n\ge 1}$ is a Cauchy sequence in $\mathbb{R}$.)

Now, if $({f}_{n})$ is any sequence of step functions, then $\{x\in X:|{f}_{m}(x)-{f}_{n}(x)|<\epsilon \}$ 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.

If $({f}_{n})$ is a sequence of real-valued functions on a set $X$, then its point of convergence is given by

$E=\bigcap _{\epsilon \in {\mathbb{Q}}_{>0}}\bigcup _{N\ge 1}\bigcap _{m,n\ge N}\{x\in X:|{f}_{m}(x)-{f}_{n}(x)|<\epsilon \}.$

(Note that $E$ is precisely the set of all $x$ at which $({f}_{n}(x){)}_{n\ge 1}$ is a Cauchy sequence in $\mathbb{R}$.)

Now, if $({f}_{n})$ is any sequence of step functions, then $\{x\in X:|{f}_{m}(x)-{f}_{n}(x)|<\epsilon \}$ 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.

Micaela Simon

Beginner2022-06-28Added 3 answers

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.

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