What is the reason this example of an everywhere surjective function have or does not have an undefined mean?
Suppose $f:\mathbb{R}\to\mathbb{R}$ is everywhere surjective (i.e., $f[(a,b)]=\mathbb{R}$ for all non-empty intervals $(a,b)$) where the graph of $f$ has Hausdorff dimension $2$ with zero $2$-d Hausdorff measure—i.e., the measure is defined on the Borel $\sigma$-algebra.
Here is an example.
Question: For each $c,d\in\mathbb{R}$, is the mean of $\mathsf{F}:=\left. f\right|_{(c,d)}$ (Definition $1$) undefined? If so, what is the reason? If not, then what is the reason?
Definition $\S$1. (Mean of $\mathsf{F}$)
Suppose:
- $\dim_{\text{H}}(\cdot)$ is the Hausdorff dimension
- $\mathcal{H}^{\dim_{\text{H}}(\cdot)}(\cdot)$ is the Hausdorff measure in its dimension on the Borel $\sigma$-algebra.
- the integral of $\mathsf{F}$ is defined w.r.t the Hausdorff measure in its dimension
The expected value of $\mathsf{F}:A\to\mathbb{R}$ (i.e., $A:=(c,d)$), w.r.t. the Hausdorff measure in its dimension, is $m_{{}_{\mathsf{F}}\!}(A)$ (when it exists) where:
$$m_{{}_{{\large{\mathsf{F}}}}\!}(A)=\frac{1}{{\mathcal{H}}^{\text{dim}_{\text{H}}(A)}(A)}\int_{A}\mathsf{F}\, d{\mathcal{H}}^{\text{dim}_{\text{H}}(A)}$$
2 answers
Answer from u/kuromajutsushi on Reddit:
Before anyone wastes their time on this:
You already asked this question, and I already answered it. In the example you link to in your question, let $h(x)$ be this function, which is a continuous function whose graph has Hausdorff dimension $2$. Then on each interval $(c,d)$, $g(x)$ is equal to a continuous function a.e. so its integral is finite.
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Answer from math stack exchange (I looked up the user online and I am not sure he specializes in mathematics; therefore, take his answer with a pinch of salt):
Yes, the other user is right. The expected value is not undefined because the graph has Hausdorff dimension 2 or zero 2-D Hausdorff measure; that’s irrelevant since expectation integrates over the domain, not the graph. The real reason is that an everywhere-surjective function on every subinterval must be wildly pathological and cannot be Lebesgue-integrable, so the integral defining the expectation fails to exist. The geometry of the graph doesn’t determine integrability; it’s the function’s behavior as a measurable function on (c,d) that matters.
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