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I am trying to prove that if $Z$ is a zero set and $\omega$ is an abstract outer measure, and $E$ and $X$ are sets, then

$\omega(X \cap E^c \cap Z^c)= \omega(X \cap E^c )$

This intuitively makes sense because $\omega(Z^c)$ should equal $\omega(M)$, where $M$ is the universe. So intersecting a set with $Z^c$ shouldn't affect the measure of that set. I'm not sure how to prove this though.

For context, this is part of a larger proof that if $E$ is measurable, then $E \cup Z$ is measurable.

Bernard
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Vladimir Lenin
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    Is a "zero set" any set of measure zero? – David K May 01 '21 at 20:44
  • In the construction of a measure out of an outer measure, any $A$ of outer measure $0$ are measurable in the sense of Caratheodory, that is for any $E\subset X$, $m^(E)=m^(E\cap A)+m^(E\setminus A)$. Of course, as $m^(A\cap E)=0$ if $A$ has outer measure $0$. Review the construction of measures out of outer measures. For example here and here – Mittens May 02 '21 at 01:42
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    Note: A set of zero measure is a null set; otoh a "zero set" is usually ${x:f(x)=0}$ for some function $f$. – David C. Ullrich May 02 '21 at 13:29

1 Answers1

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Let us prove:

Let $\omega$ be an abstract outer measure and $Z$ be any set such that $\omega(Z)=0$. Then for any set $F$, $\omega(F \cap Z^c) = \omega(F)$ .

Proof: By the monotonicity of $\omega$, we have that $$\omega(F \cap Z^c) \leq \omega(F) \tag{1}$$

On the other hand, by the sub-sdditiviy, we have $$ \omega(F) \leq \omega(F \cap Z) + \omega(F \cap Z^c) $$ Since $\omega(F \cap Z) \leq \omega( Z) =0$, we have that $ \omega(F \cap Z) =0$. So we have $$ \omega(F) \leq \omega(F \cap Z^c) \tag{2}$$

From $(1)$ and $(2)$, we have $\omega(F \cap Z^c) = \omega(F)$. $\square$

For the case in your question, just take $F=X \cap E^c$.

Ramiro
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