$A,B\subset\mathbb{R}, |A|<\infty\Rightarrow |B-A|\geq |B|-|A|,\ |\cdot|$ outer measure

measure-theoryouter-measurereal-analysissolution-verification

I have proved the following statement and I would like to know if I have made any mistakes, thanks.

"$A,B\subset\mathbb{R}, |A|<\infty\Rightarrow |B-A|\geq |B|-|A|$"

where $|\cdot|$ denotes outer measure.

My proof:

(1) countable subadditivity of outer measure; (2) outer measure preserves order

(I) $$|B|=|(B-A)\cup (A\cap B)|\overset{(1)}{\leq}|B-A|+|A\cap B|\Rightarrow |B-A|\geq |B|-|A\cap B|$$

(II) $$A\cap B\subset A\overset{(2)}{\Rightarrow} |A\cap B|\leq |A|\overset{\text{hyp.}}{<}\infty \Rightarrow -|A\cap B|\geq -|A|\overset{*}{\Rightarrow} |B|-|A\cap B|\geq |B|-|A|\overset{(I)}{\Rightarrow} \fbox{$|B-A|\geq |B|-|A|$}$$

*note that the following inequality is valid because $|A\cap B|<\infty$ otherwise we could have things like $\infty-\infty$ if $|B|=\infty$ or, if $|B|<\infty$ too, $-\infty\geq |B|-|A|\in\mathbb{R}$, absurd

Best Answer

After your edits, your solution looks pretty good and I have little in the ways of mathematical criticism to offer. I would just say that it is generally considered better style to write more words, fewer symbols and avoid "itemizing" the logical structure of your proof more than necessary. Basically, you want your proof to look more like an ordinary piece of prose, and less like computer code.

A solution more in keeping with these principles might look something like this:

Because $B \subseteq B \cup A = (B-A) \cup A$, we have $$|B| \leq |(B-A) \cup A| \leq |B -A| + |A|.$$ The first inequality above uses monotonicity of outer measure and the second uses subadditivity. Because $|A|<\infty$, we may subtract it on both sides to obtain the desired inequality: $$|B| - |A| \leq |B-A|.$$

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