For $A \subseteq \mathbb{R}$. We know that $m^*(A + t) = m^*(A)$. but Does it follow that
$$ m^*(\lambda A) = \lambda m^*(A)$$??? for $\lambda > 0$. Here $m*: P( \mathbb{R}) \to [0, \infty ] $ is lebesgue outer measure .
For $A \subseteq \mathbb{R}$. We know that $m^*(A + t) = m^*(A)$. but Does it follow that
$$ m^*(\lambda A) = \lambda m^*(A)$$??? for $\lambda > 0$. Here $m*: P( \mathbb{R}) \to [0, \infty ] $ is lebesgue outer measure .
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It doesn't follow directly. It follows from that $\{I_j\}_{j=1}^{\infty}$ covers $A$ if and only if $\{\lambda I_j\}_{j=1}^{\infty}$ covers $\lambda A$.