This exercise comes from: Sheldon Axler's "Measure, Integration, and Real Analysis". Section 2A, Exercise 2.
I have read the answers to the following, and understand the proofs.
(1)$A\subset\mathbb{R}, t\in\mathbb{R}\Rightarrow |tA|=|t||A|$
(2)Analysis question on outer measure
Nevertheless I cannot figure out why the following reasoning is wrong:
(Note: $tA$ is defined as $\{ta | a \in A\}$)
Finite sets have outer measure $0$. Since $t$ is a single element, $t$ has outer measure $0$. Therefore, $$|t||A| = 0 |A| = 0.$$ for all $t \in \mathbb{R}$ and all $A \in \mathbb{R}$.
Yet, in the case that $t = 1$, with $A$ being any open interval of $\mathbb{R}$, we have that
$$|tA| = \{1a | a \in A\} = |A|$$
Thus,
$$ 0 = |t||A| \neq |tA| = A $$
So the statement that $|t||A| = |tA|$ is false.
Why am I wrong? Thank you.
$|t|$ is absolute value of $t\in\Bbb{R}$ . Can you prove that $|t|=\lambda^{*}(\{t\})$?
$\lambda^{*}$ : Lebesgue outer measure.
$\lambda^{*}(tA) =|t|\lambda^{*}(A)$
no danger of confusion.