Can two sets $A,A'\subset\mathbb{R}^n$ have the same measure but different expectation? 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:

*

*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., $\mathbb{E}[A]\neq\mathbb{E}[A’]$)?


*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’] $)?


*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.
 A: 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:

*

*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'] $.


*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 $0$.


*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 $0$).
