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I have three 'simple' questions regarding measure theory. For every measurable set $A\subset\mathbb{R}^n$, we define the avarage value of $A$ (denoted as $\mathbb{E}[A]$) by \begin{gather} \mathbb{E}[A]=\frac{1}{\mu(A)}\int_A \vec{x} \mathrm{d}\mu(\vec{x}) \end{gather} where $\mu$ is any measure defined on the Borel $\sigma$-algebra of $\mathbb{R}^n$.

My three questions are as follows:

  1. Can two sets $A,A'\subset\mathbb{R}^n$ such that $A\neq A'$ have the same measure (i.e., $\mu(A)=\mu(A’)$) but different average values (i.e., $\mathbb{E}[A]\neq\mathbb{E}[A’]$)?

  2. When will two sets $A,A'\subset\mathbb{R}^n$ such that $A\neq A'$ have the same avarage value (i.e., $\mathbb{E}[A]=\mathbb{E}[A’] $)?

  3. Will any set $A\subset\mathbb{R}^n$ with measure $\mu(A)=\infty$ have avarage value $\mathbb{E}[A]=\infty$?

My attempt to reasoning about these questions is the following.

For question 1, consider the Real line $\mathbb{R}$ and sets $A=[1,2]$ and $A'=[3,4]$. Since the notion of measure in the Real line boils down to the concept of length, the measure $\mu$ of each set is just $\mu(A)=2-1=1$ and $\mu(A')=4-3=1$. However, the average value of each set is just $\mathbb{E}[g_A]=1.5$ and $\mathbb{E}[g_{A'}]=3.5$. Therefore, for the case in which $n=1$, it is possible to construct two sets $A,A'\subset\mathbb{R}$ with the same measure but different average values. Hence, I'm inclined to believe that the answer to question 1 is 'yes', but I do not know if this argument generalises beyond $n=1$.

For question 2, I do not have a clear argument, but I guess that two sets $A,A'\subset\mathbb{R}^n$ will have the same average value when their measure is zero (i.e., $\mu(A)=\mu(A')=0$), when their measure is infinite (i.e., $\mu(A)=\mu(A')=\infty$) or when they are not disjoint and the only sets that receive strictly positive measure belong to the intersection $A\cap A'$. Am I correct?

For question 3, I do not have an argument... My intuition just tells me that the answer is 'yes'.

As you can probably tell from my questions, I don't quite understand what does it mean for a set to have a certain mesure. Any explanation will be very welcome.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ Commented Mar 8, 2021 at 23:58

1 Answer 1

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Let us start presenting some "intuition". A measure $\mu$ can be thought as way to determine the sizes of sets. The average value a set $A$ with respect to $\mu$ is the center of $A$ if we measure sizes using $\mu$.

Now for your questions:

  1. Your argument is correct and valid in $\Bbb R^n$. Just an example in $\Bbb R^2$: $A= [1,2]\times [0,1]$ and $A'= [3,4]\times [0,1]$. Let us use Lebesgue measure in $\Bbb R^2$. Then $\mu(A)=\mu(A') = 1$, but for the average values, we have: $\mathbb{E}[A] =(1.5,0.5)\neq (3.5,0.5) =\mathbb{E}[A'] $.

  2. The are many situation were distinct sets can have the same average value. For instance:

a. $A = [-1, 1]$ and $A'=[-100, 100]$. Let us use Lebesgue measure in $\Bbb R$. They are distinct sets with different "sizes" but the same average value, that is $0$. Note that there are subsets of $A'\setminus A$ that have strictly positive measure.

b. $A = [-1, 1]$ and and $A'=[-100, -2] \cup [2,100]$. Let us use Lebesgue measure in $\Bbb R$. They are disjoint sets with different "sizes" but the same average value, that is $0$.

  1. If $\mu(A) = \infty$ then, in general, the average point is not defined. The formula \begin{gather} \mathbb{E}[A]=\frac{1}{\mu(A)}\int_A \vec{x} \mathrm{d}\mu(\vec{x}) \end{gather} will produced an undetermined value similar to $\frac{\infty}{\infty}$ in the case of $\Bbb R$.

Remark 1: There are some cases where $\mu(A)=\infty$ and the average value is defined. For instance, consider $\Bbb R$ and let $\mu$ be a measure defined by $\mu(E)=\infty$ if $0 \in E$ and $\mu(E)=0$ if $0 \notin E$. Let $A =[-1,1]$. In such case, $\mu(A) = \infty$ and $$ \int_A x \mathrm{d}\mu(x) = 0$$ So \begin{gather} \mathbb{E}[A]=\frac{1}{\mu(A)}\int_A x \mathrm{d}\mu(x) = 0 \end{gather}

Remark 2: Using the formula \begin{gather} \mathbb{E}[A]=\frac{1}{\mu(A)}\int_A \vec{x} \mathrm{d}\mu(\vec{x}) \end{gather} It is easy to see that, if $\mu(A)=\infty$ and the average value is defined, then necessarily, the average value is $0$.

Remark 3: There are some ways to extend the notion of average value to broader classes of sets of infinite measure, but which way is suitable depends on the context (for instance, using limits of integral, in a certain way, we could say that, for the Lebesgue measure, the average value of $\Bbb R$ is $0$).

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  • $\begingroup$ Thank you very much for your insightful answer. I am now trying to fully grasp the details. I will accept your answer in a while (perhaps with the addition of some comment asking for clarification). $\endgroup$ Commented Mar 8, 2021 at 13:27
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    $\begingroup$ Some technical points on (3): First, whether the average point is defined depends on both the measure and the set, but the only value it could be is zero when it's defined (so you may have been forgetting this as an edge case, e.g. counting measure and a countable set whose sum is absolutely convergent). Second, if we use the improper Riemann integral, the average value of $\mathbb{R}$ is still undefined. It's possible you meant the Cauchy p.v. $\endgroup$ Commented Mar 8, 2021 at 17:11
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    $\begingroup$ @BrianMoehring I tried to keep things simple for the example. I will try to add details without making the answer too complex. $\endgroup$ Commented Mar 8, 2021 at 17:23
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    $\begingroup$ While I'm definitely sympathetic to Keeping It Simple, it's a common enough error (that the improper Riemann integral of odd continuous functions is zero) that I'd contend the only simple+harmless explanation keeps the integral vague. $\endgroup$ Commented Mar 8, 2021 at 17:34
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    $\begingroup$ @BrianMoehring You are right. Improper Riemann integral of odd continuous functions is not zero. To avoid inducing the incorrect message, I will change the wording. $\endgroup$ Commented Mar 8, 2021 at 17:46

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