Is the Lebesgue measure of $\mathbb R$ (reals) infinity? if so, can somebody tell me the Lebesgue measure of $[0,1] \cup \mathbb Q$ (rationals)? Just a curiosity question.
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$\begingroup$Countable additivity and translation invariance answer both questions if you know that the measure of $[0,1]$ is $1$ and the measure of a set containing just one point is $0.$ The measure of every set of the form $[a,a+1]$ must be the same as the measure of $[0,1]$ by translation invariance. The whole line contains the union of a countably infinite collection of such sets that are pairwise disjoint; for example $\displaystyle\bigcup_{\text{even } n} [n,n+1].$ Hence the measure of the real line is $\ge1+1+1+\cdots = \infty$ and is therefore $\infty.$
The set of all rational numbers is countable, so its measure is the sum of the measures of sets each containing just one rational number. That means it is $0+0+0+\cdots = 0.$
Therefore the measure of $[0,1]\cup\mathbb Q$ is $1.$
$\endgroup$ 2 $\begingroup$If $A$ and $B$ are measurable sets then so are $A\cup B, B\cap A, A$ \ $B,$ and $B$ \ $A.$ Therefore $$m(A)\leq m(A)+m(B \backslash A)\leq m(A)+[m(B \backslash A)+m(B\cap A)]=m(A)+m(B).$$ Since $m(A)+m(B$ \ $A)=m(A \cup (B$ \ $A))=m(A\cup B)$ we have $$ m(A)\leq m(A\cup B)\leq m(A)+m(B).$$ In particular if $m(B)=0$ then $m(A)=m(A\cup B).$
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