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$$\mu(E_1\cup E_2) = \sup\{\Lambda f: f\prec E_1\cup E_2\}$$ For every $f\prec E_1\cup E_2$, define $f_1 = f\vert_{E_1}$ on $E_1$ and $0$ otherwise. Similarly, define $f_2 = f\vert_{E_2}$ on $E_2$ and $0$ otherwise. Then, $f = f_1 + f_2$. Note that $f_1 \prec E_1$ and $f_2\prec E_2$. $$\mu(E_1\cup E_2) = \sup\{\Lambda f_1 + \Lambda f_2: f_1\prec E_1, f_2\prec E_2\} \\ = \sup\{\Lambda f_1: f_1\prec E_1\} + \sup\{\Lambda f_2: f_2\prec E_2\} = \mu(E_1) + \mu(E_2)$$
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Wed, 19 May 2021 16:36 GMT