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!
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.