Interactive Real Analysis - part of MathCS.org

• Real Analysis
• 1. Sets and Relations
• 2. Infinity and Induction
• 3. Sequences of Numbers
• 4. Series of Numbers
• 5. Topology
• 6. Limits, Continuity, and Differentiation
• 7. The Integral
• 7.1. Riemann Integral
• 7.2. Integration Techniques
• 7.3. Measures
• 7.4. Lebesgue Integral
• 7.5. Riemann versus Lebesgue
• 8. Sequences of Functions
• 9. Historical Tidbits
• Java Tools

Examples 7.3.7(c): Measurable Sets

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Show that the union of two measurable sets is measurable.

Context

Assume that E and F are two measurable sets. We need to prove that for every set A we have:

m^*(A) m^*(A (E F) + m^*(A comp(E F))

We know that

1. E is measurable so that for every set A we have:

   m^*(A) = m^*(A E) + m^*(A comp(E))

2. F is measurable so that for every set A we have:

   m^*(A) = m^*(A F) + m^*(A comp(F))

3. From set theory we know (draw a Venn diagram to verify) that:

   A (E F) = (A E) (A comp(E) F)

   which implies by subadditivity of m^* that

   m^*(A (E F)) m^*(A E)) + m^*(A comp(E) F))

Using A comp(E) in place of A in (2) gives:

m^*(A comp(E)) =
      = m^*(A comp(E) F) + m^*(A comp(E) comp(F)) =
      = m^*(A comp(E) F) + m^*(A comp(E F))

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

m^*(A) = m^*(A E) + m^*(A comp(E) F) + m^*(A comp(E F))
      m^*(A (E F)) + m^*(A comp(E 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).