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Let $\mu$ be a Lebesgue measure on the Borel $\sigma$ algebra.

Then is $\mu( [0,\frac{1}{4}) \bigcup [\frac{3}{4},1])$ just $\mu([\frac{3}{4},1]) + \mu([\frac{3}{4},1])$

with $\mu( [0,\frac{1}{4})) = \frac{1}{4} - 0 = \frac{1}{4}$

and $\mu([\frac{3}{4},1]) = 1 - \frac{3}{4} = \frac{1}{4}$?

Ethan Bolker
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josh
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1 Answers1

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Your calculations are correct, since the two intervals $[0, 1/4)$ and $[3/4, 1]$ are disjoint and $\mu(A \cup B) = \mu(A) + \mu(B)$ for any disjoint measurable sets $A, B$ and measures $\mu$.

Dominik
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