0

In A. J. Weir’s Lebesgue Integration and Measure (CUP 1973) the author proves that, given an increasing sequence of step functions $\phi_n$ for which the sequence $\smallint \phi_n$ converges, the points $x$ for which the sequence $\phi_n(x)$ diverge comprise a null set (or, a set of measure zero).

Subsequently Weir proves a ‘converse’ whereby, given a null set $S$ an increasing sequence of step functions with bounded (and therefore convergent) integrals is constructed, say $\phi_n$, such that, for every $s \in S$ the sequence $\phi_n(s)$ diverges.

Weir then says that this property ‘characterises’ null sets.

But the proof doesn’t show this because the constructed $\phi_n$ don’t converge pointwise for elements not in S.

Can anyone complete the proposed characterisation, or prove it false?

Steve Powell
  • 192
  • 1
  • 8
  • As an exercise I used the proof to construct a sequence of step functions that applied to the rationals — a well-known null set since it is countable. – Steve Powell Jun 21 '20 at 15:26
  • "Step function" here means something of the form $\phi = \sum_{j=1}^m a_j 1_{I_j}$ where $I_j$ are (open, closed, half-open) intervals? – Nate Eldredge Jun 21 '20 at 15:26
  • @NateEldredge If $1_{I_j}$ denotes the characteristic function on the interval $I_j$, then yes, that is exactly what I mean. Weir develops the Lebesgue integral before going on to define the measure, and he does so by taking limits of step functions to generate the integrable functions. – Steve Powell Jun 21 '20 at 15:31
  • If so then it seems that the set where the $\phi_n$ diverge must be $\mathbf{\Pi}^0_3$, aka $F_{\sigma \delta}$, and that is certainly not true of every null set. – Nate Eldredge Jun 21 '20 at 15:31
  • In particular the set where the $\phi_n$ diverge is certainly Borel, so any non-Borel null set will be an easy counterexample. – Nate Eldredge Jun 21 '20 at 15:36
  • @NateEldredge Hmm; I’m not sure what is meant by these notations, but the first theorem I quoted certainly is true, so I’m not sure what you are claiming. Could you answer the question in the negative, then? – Steve Powell Jun 21 '20 at 15:37
  • @NateEldredge The Borel angle sounds like a better approach. I’ll research it. – Steve Powell Jun 21 '20 at 15:40

1 Answers1

1

Each of the functions $\phi_n$ is Borel, therefore the set $A$ of all points $x$ such that $\phi_n(x)$ diverges is a Borel set (standard exercise). But there exist null sets $S$ that are not Borel; for instance, all of the $2^{\mathbb{c}}$ subsets of the Cantor set are null, but there are only $\mathfrak{c}$ many Borel sets in total. See also Is there a null set that is not a Borel set?, Lebesgue measurable but not Borel measurable. So if we take $S$ to be such a set, then there cannot exist any sequence of step functions $\phi_n$, nor indeed any sequence of Borel functions, such that $\phi_n(x)$ diverges iff $x \in S$.

In other words, if $S$ is such a set, Weir's result will show you can find an increasing sequence of step functions $\phi_n$ such that $\phi_n(x)$ diverges for every $x \in S$ and $\int \phi_n$ converges, but there will always also be some $x \notin S$ (indeed, uncountably many) for which $\phi_n(x)$ diverges as well.

With a little more care, one can show that for step functions $\phi_n$, the set $A$ is necessarily a countable intersection of countable unions of closed sets, denoted $F_{\sigma \delta}$ or $\mathbf{\Pi}^0_3$ in the Borel hierarchy. Since the Borel hierarchy doesn't collapse, I think this shows there are Borel subsets of the Cantor set which are not $F_{\sigma \delta}$ and therefore one could find Borel counterexamples as well.

I think Weir's terminology "characterization" is not wrong, in that it does give a necessary and sufficient condition for a set to be null, but it has to be interpreted carefully. The following statement is what he proved:

A set $S$ is null if and only if there exists an increasing sequence of step functions $\phi_n$, with $\int \phi_n$ convergent, such that $x \in S \implies \phi_n(x)\text{ diverges}$.

You seem to be thinking of the following different statement, which is false:

(FALSE) A set $S$ is null if and only if there exists an increasing sequence of step functions $\phi_n$, with $\int \phi_n$ convergent, such that $x \in S \Longleftrightarrow \phi_n(x)\text{ diverges}$.

Nate Eldredge
  • 101,664
  • Thank you Nate, you have resolved a question that has bugged me for years. I can finally update my personal notes in my copy of Weir’s book! – Steve Powell Jun 21 '20 at 15:49
  • I still think the term characterisation is too strong a word, but I see the distinction in the two statements and reluctantly agree that that is what Weir might have meant. P.S. in your statement you omitted the convergence condition on the increasing sequence of step functions. – Steve Powell Jul 02 '20 at 16:18
  • @StevePowell: I have corrected this, thanks. – Nate Eldredge Jul 02 '20 at 16:43