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
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Proposition 7.3.11: Equivalent Measurable Sets

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Let f be a real-valued function with a measurable set as domain. Then the following are quivalent:

1. the set { f(x) a } is measurable for all real numbers a
2. the set { f(x) < a } is measurable for all real numbers a
3. the set { f(x) a } is measurable for all real numbers a
4. the set { f(x) > a } is measurable for all real numbers a

Context

The proof is not too bad. Remember, we do not need to prove that any of these sets are measurable per se; we need to prove that if one of them is measurable then the others are, too.

So, assume that (1) is true. Then (4) must be true since it is the complement of (1), and we know that complements of measurable sets are measurable. This clearly works the other way around as well, so that (1) is equivalent to (4).

Again by considering complements we can see that (2) is equivelent to (3).

Finally we can explore the fact that unions and intersections of measurable sets are measurable together with:

{ f(x) < a } = _{n > 0} { f(x) a - 1/n }

to see that (1) implies (2). Similarly because:

{ f(x) a } = _{n > 0} { f(x) < a + 1/n }

we see that (2) implies (1). But then our proof is finished.