description/proof of that for measure space and measurable subset, Lebesgue integral of integrable complex function over space is integral over subset plus integral over complement of subset
Topics
About: measure space
The table of contents of this article
Starting Context
- The reader knows a definition of Lebesgue integral of integrable complex function over measurable subset of measure space.
- The reader admits the proposition that for any measure space and for any measurable subset, the Lebesgue integral of any measurable extended real function over the space is the integral over the subset plus the integral over the complement of the subset.
Target Context
- The reader will have a description and a proof of the proposition that for any measure space and for any measurable subset, the Lebesgue integral of any integrable complex function over the space is the integral over the subset plus the integral over the complement of the subset.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\((M, A, \mu)\): \(\in \{\text{ the measure spaces }\}\)
\( \mathbb{C}\): \(= \text{ the complex Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(f\): \(: M \to \mathbb{C}\), \(\in \{\text{ the Lebesgue integrable functions }\}\)
\(a\): \(\in A\)
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Statements:
\(\int_M f d \mu = \int_a f d \mu + \int_{M \setminus a} f d \mu\)
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2: Proof
Whole Strategy: Step 1: see \(f\) is integrable over \(a\) and over \(M \setminus a\); Step 2: apply the proposition that for any measure space and for any measurable subset, the Lebesgue integral of any measurable extended real function over the space is the integral over the subset plus the integral over the complement of the subset.
Step 1:
\(f = Re (f) + i Im (f) = Re (f)^+ - Re (f)^- + i (Im (f)^+ - Im (f)^-)\).
\(f\)'s being integrable means that \(\int_M Re (f)^+ d \mu \lt \infty\), \(\int_M Re (f)^- d \mu \lt \infty\), \(\int_M Im (f)^+ d \mu \lt \infty\), and \(\int_M Im (f)^- d \mu \lt \infty\).
Then, \(\int_a Re (f)^+ d \mu \lt \infty\), \(\int_a Re (f)^- d \mu \lt \infty\), \(\int_a Im (f)^+ d \mu \lt \infty\), and \(\int_a Im (f)^- d \mu \lt \infty\), because \(Re (f)^+\), \(Re (f)^-\), \(Im (f)^+\), and \(Im (f)^-\) are non-negative.
So, \(f\) is integrable over \(a\).
\(f\) is integrable over \(M \setminus a\), likewise.
Step 2:
\(\int_M f d \mu = \int_M Re (f) d \mu + i \int_M Im (f) d \mu\).
\(= \int_a Re (f) d \mu + \int_{M \setminus a} Re (f) d \mu + i (\int_a Im (f) d \mu + \int_{M \setminus a} Im (f) d \mu)\), by the proposition that for any measure space and for any measurable subset, the Lebesgue integral of any measurable extended real function over the space is the integral over the subset plus the integral over the complement of the subset, \(= \int_a Re (f) d \mu + i \int_a Im (f) d \mu + \int_{M \setminus a} Re (f) d \mu + i \int_{M \setminus a} Im (f) d \mu = \int_a f d \mu + \int_{M \setminus a} f d \mu\).
