If $A_1 \subset A_2 \subset \mathbb R$ and $m^*(A_1) = m^*(A_2)$, will $m^*(A_1 \cap T) = m^*(A_2 \cap T), \forall T \subset \mathbb R$?

Question

Definition of Lebesgue Outer Measure: Given a set $E$ of $\mathbb R$, we define the Lebesgue Outer Measure of $E$ by, $$m^*(E) = \inf \left\{\sum_{n=1}^{+\infty} l(I_n): E \subset \bigcup_{n=1}^{+\infty}I_n \right\}$$ where $l(I_n)$ denotes the length of interval (bounded and nonempty interval).

Definition of measurable set: A set $E$ measurable if $$m^*(T) = m^*(T \cap E) + m^*(T \cap E^c)$$ for every subset of $T$ of $\mathbb R$.

If $A_1 \subset A_2 \subset \mathbb R$ and $m^*(A_1) = m^*(A_2)$, will $$m^*(A_1 \cap T) = m^*(A_2 \cap T)$$ for all $T \subset \mathbb R$? Why? How about the result if $A_1$ being a measurable set added?

Update:

Sincerely, I appreciate user140776's answer and he gives the counterexample that equality doesn't hold for $m^*(A_1) = m^*(A_2) = +\infty$. I still have a question: if $m^*(A_1) = m^*(A_2) < +\infty$, does the equality hold? Or it doesn't hold until adding the condition that $A_1$ is measurable?

Besides, if I remove the restriction of $A_1$ being a subset of $A_2$ that is $A_1, A_2 \subset \mathbb R$ and $m^*(A_1) = m^*(A_2) < +\infty$, will that equality $$m^*(A_1 \cap T) = m^*(A_2 \cap T)$$ for all $T \subset \mathbb R$ still hold?

Answer 1

I think the answer is no.

Counterexample: Let $A_1=(1,\infty)$, $A_2=(0,\infty)$ and $T=(0,1)$. Then $A_1\subset A_2\subset \mathbb{R}$, $m^*(A_2)=m^*(A_1)=\infty$ and $A_1$ is measurable. However, $m^*(A_1\cap T)=0 \ne 1=m^*(A_2 \cap T)$.

However, if in addition to $A_1$ being measurable, you also assume that $m^*(A_1)<\infty$, then I think maybe the result will follow from the Caratheodory condition, but I'm not 100% sure.

Answer 2

Here's a strong counterexample: For every $X \subseteq [0, 1]$, there is a $Y \subseteq X$ such that $\mu^{\star}(X) = \mu^{\star}(Y) = \mu^{\star}(X \setminus Y)$. So letting $A_1 = Y, A_2 = X, T = X \setminus Y$ gives you a strong counterexample. For our purpose, it is enough to take $A_2 = [0, 1]$ and $A_1$ to be a Bernstein subset of $[0, 1]$ (which means that both $A_1, [0, 1] \setminus A_1$ meet every perfect subset of $[0, 1]$).