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There is already a similar question here on MSE except the op asked for an increasing sequence. This is quickly seen to be false, so I'm asking about a sequence which is not necessarily increasing. My intuition seems to be that there is no such sequence and I've tried to prove it but failed. The argument I was think of is that given a particular $x$ in $Q$ if a step function converges to it then for an irrational point "near enough" $x$ it shouldn't; however, since pointwise convergence depends on the point x (i.e. finding N depends on x) and the sequence of step functions isn't necessarily increasing, I got stuck in attempting to prove it. Perhaps my intuition is wrong and I would appreciate help answering this question.

Leonid
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    There is a theorem in Stein & Shakarchi that says for measurable $f:\mathbb{R}^d \rightarrow \mathbb{R}$, there is a sequence ${\psi_n}_n$ of step functions such that $\psi_n \rightarrow f$ almost everywhere. I think that should give it to you, unless you want true point wise convergence. – Joe Jul 07 '22 at 18:42
  • I know of this theorem. Which is why I'm asking about pointwise convergence everywhere not almost everywhere. The theorem can't be adapted to this case because the key idea in that proof is that you can approximate arbitrary measurable sets by finite union of intervals measure-wise which helps you disregard complicated sets of measure zero. – Leonid Jul 07 '22 at 18:47
  • Also, If I were talking about a.e. convergence then taking the constant 0 sequence would suffice which is obviously not what I'm looking for. – Leonid Jul 07 '22 at 18:52
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    You might want to look here: https://math.stackexchange.com/questions/4311105/a-measurable-function-that-does-not-admit-pointwise-approximation-by-step-functi?rq=1

    This answer seems to indicate that every set sufficiently low in the Borel Hierarchy has a characteristic function that is the limit of step functions. Since $\mathbb{Q}$ is a $\mathbf{\Sigma}_2^0$ set, you should be able to get such a sequence of step functions. I may think about this more later.

     – Joe Jul 07 '22 at 19:14
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    @coffeemath I was talking about a.e. convergence. My point was that constructing a sequence of simple function converging a.e. to the Dirichlet function is trivial since the Dirichlet function is already equal a.e. to a simple function, namely the zero function. – Leonid Jul 07 '22 at 20:10
  • Got it/ deleting comment. – coffeemath Jul 07 '22 at 21:00

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I have an answer but you’re not going to like it. Using Stein & Shakarchi’s definitions, a step function is a function $S:\mathbb{R}^n \rightarrow \mathbb{R}$ of the form: $$S(x) = \sum_{j=1}^n \chi_{R_j}(x)$$ Where $\chi_A$ denotes the characteristic function of the set $A$, and $R_j$ is a rectangle for each $j$. The obvious question is “what exactly do Stein and Shakarchi mean by ‘rectangles’?”

On page 3, they write:

A (closed) rectangle $R$ in $\mathbb{R}^d$ is the product of $d$ one-dimensional closed and bounded intervals $$R = [a_1, b_1] \times \cdots \times [a_d, b_d],$$ where $a_j \leq b_j$ are real numbers, $j=1,2, \dots d$.

Since we allow $a_j = b_j$, it follows that a singleton is a rectangle. Thus we may trivially take a sequence of step functions that converge pointwise to the Dirichlet function by taking: $$S_n (x) := \sum_{j=1}^n \chi_{\{q_j\}}(x)$$ Where $\{q_n\}_{n \in \omega}$ is an enumeration of $\mathbb{Q}$.

I imagine you have already thought of this, and would like to avoid such “trivial” step functions.

In my comment, I mentioned a stack exchange post that indicates that for a set $B \subseteq \mathbb{R}$, the characteristic function $\chi_B$ is the pointwise limit of step functions if and only if $B$ is a $\mathbf{\Pi}^0_4$. I’m not exactly sure if this is true, but I imagine that the tools in Kechris’s book on Descriptive Set Theory may be helpful.

Assuming that result is true (big if), I’m curious if the result relies on these kinds of “trivial” step functions, or if they can be avoided.

Joe
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  • Yes I was hoping to avoid degenerate intervals. Your other comment seems to be helpful though. But in the link you posted I didn't find the claim where a characteristic function of a measurable set is a pointwise limit of step functions if and only if it belongs to a certain class of the Borel hierarchy. – Leonid Jul 08 '22 at 10:58
  • Like I said, they don’t explicitly say that, but I think you could derive it. My descriptive set theory skills are a bit rusty, but it’s an interesting problem and I may return to it later. If I do, I will write more in my answer. – Joe Jul 08 '22 at 16:02