# A basic question on the definition of probability measure

Let $\Omega$ denote the unit interval $(0,1]$. Let $\omega$ denote the generic point of $\Omega$. Denote the length of an interval $I =(a,b]$ by $|I|$ : |I|=b-a

If $A=\cup_{i=1}^{n}I_i= \cup_{i=1}^{n}(a_i,b_i]$ where the intervals $I_i$ are disjoint and contained in $\Omega$ assign to $A$ the probability $P(A)=\sum_{i=1}^{n}|I_i|=\sum_{i=1}^{n}(b_i-a_i)$.

If $A$ and $B$ are two such finite disjoint unions of intervals, and if $A$ and $B$ are disjoint then $A \cup B$ is a finite disjoint union of intervals and $P(A \cup B)=P(A) + P(B)$. I think this follows easily from the definition of $P(A)$. But, I have read in a book that this is a consequence of the additivity of Riemann integral. Do we need that ?

In any case, yes, your "measure" does satisfy that $P(A\bigcup B)=P(A)+P(B)$. You could also make it a consequence of Riemann integral by considering $\int_{0}^{1}\chi_{A}dx+\int_{0}^{1}\chi_{B}dx$. They have only finite number of discontinuity, so they are indeed Riemann integrable.