Here is an excercise of measure theory and real analysis. can you help me with it? Let (X,M, &#

Davion Harding

Davion Harding

Answered question

2022-06-23

Here is an excercise of measure theory and real analysis. can you help me with it?
Let (X,M, μ) be a measure space and f : X [ 0 , ] a measurable function. Prove that
X f d μ > 0 μ ( { x X f ( x ) > 0 } ) > 0.
I tried to use the definition of the integral by the approximation theorem of measurable functions by simple functions, but I haven't been able to come to the conclusion.

Answer & Explanation

Jake Mcpherson

Jake Mcpherson

Beginner2022-06-24Added 23 answers

Let A = { x f ( x ) > 0 } and A n = { x f ( x ) > 1 n }. Note that A = n = 1 A n . If μ ( A ) > 0, then μ ( A n ) > 0 for some n. Therefore f A n f 1 n μ ( A n ) > 0.
Conversely, suppose that f > 0. Denote B = { x f ( x ) = 0 }. We have that f = A f + B f = A f. If μ ( A ) = 0, then A f = 0 and we arrive a contradiction. Therefore μ ( A ) > 0 (By definition, A f = 1 A f and the integrand 1 A f = 0 a.e. if μ ( A ) = 0. I hope that you would not ask me to prove that g = 0 whenever g = 0 a.e.)

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