by H. Donag
Last Updated October 13, 2017 21:20 PM

We define the outer measure $\lambda^*_t$ as follows: fix $t>0$, for any subset $A \subseteq \mathbb{R}$ we have $\lambda^*_t(A) = \inf \{\sum_{k=1}^\infty l(I_k) \quad \mathrm{with} \quad I_k = \emptyset \quad \mathrm{or} \quad l(I_k) \geq t \}$ where $(I_k)_{k\in \mathbb{N}}$ is a sequence of intervals which covers the set $A$, and $l(I_k)$ is the length of the interval.

I wish to prove countable subadditivity for this function. I understand that if one sets $t=0$ we effectively have the Lebesgue outer measure, whose subadditive property I know how to prove. The subadditivity of $\lambda^*_t$ should follow from this fact. However, my next question asks me to prove that for $t'>t>0$ we have $\lambda^*_{t'}(A) \geq \lambda^*_t(A)$. I suppose this could also be gleaned from the Lebesgue outer measure. However, I would be interested in seeing an explicit construction where we prove subadditivity without appealing to Lebesgue. I am not sure how to modify the argument used to prove subadditivity for Lebesgue to this measure as the variable $t$ seem to stop my intuition dead in its tracks. I feel that my lack of intuition in analysis really shows here.

Appreciate any hints. Thanks.

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