## 7.3. Measures

### Examples 7.3.7(c): Measurable Sets

Assume that * E* and

*are two measurable sets. We need to prove that for every set*

**F***we have:*

**A**m^{*}() mA^{*}((AE) + mF^{*}(comp(AE))F

We know that

is measurable so that for every set**E**we have:**A***m*^{*}() = m**A**^{*}(**A**) + m**E**^{*}(comp(**A**))**E**is measurable so that for every set**F**we have:**A***m*^{*}() = m**A**^{*}(**A**) + m**F**^{*}(comp(**A**))**F**- From set theory we know (draw a Venn diagram to verify) that:
(**A****E**) = (**F****A**) (**E**comp(**A**)**E**)**F**which implies by subadditivity of

*m*that^{*}*m*^{*}((**A****E**)) m**F**^{*}(**A**)) + m**E**^{*}(comp(**A**)**E**))**F**

Using *A comp(E)*
in place of

*in (2) gives:*

**A**m^{*}(comp(A)) =E

= m^{*}(comp(A)E) + mF^{*}(comp(A) comp(E)) =F

= m^{*}(comp(A)E) + mF^{*}(comp(AE))F

We can now substitute that into (1) to get:

m^{*}() = mA^{*}(A) + mE^{*}(comp(A)E) + mF^{*}(comp(AE))F

m^{*}((AE)) + mF^{*}(comp(AE))F

Of course we used (3) to obtain the inequality. But that's what we wanted to show: proof finished (not very enlightning, but done).

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