unions of measurable sets are measurable

Evaluation Error: SyntaxError: Unexpected token '>'

at DataviewInlineApi.eval (plugin:dataview:19027:21)
at evalInContext (plugin:dataview:19028:7)
at asyncEvalInContext (plugin:dataview:19038:32)
at DataviewJSRenderer.render (plugin:dataview:19064:19)
at DataviewJSRenderer.onload (plugin:dataview:18606:14)
at DataviewJSRenderer.load (app://obsidian.md/app.js:1:1182416)
at DataviewApi.executeJs (plugin:dataview:19607:18)
at DataviewCompiler.eval (plugin:digitalgarden:10763:23)
at Generator.next (<anonymous>)
at fulfilled (plugin:digitalgarden:77:24)

Single Unions

Proposition

If $E_{1}, E_{2}$ are measurable, then $E_{1} \cup E_{2}$ are measurable

Proof

Let $A \subset R$ . Since $E_{2}$ is measurable, we know

m^{*} (A \cap E_{1}^{c}) = m^{*} (A \cap E_{1}^{c} \cap E_{2}) + m^{*} (A \cap \underset{(E_{1} \cup E_{2})^{c}}{\underset{⏟}{E_{1}^{c} \cap E_{2}^{c}}})

Then we also know

\begin{aligned} A \cap (E_{1} \cup E_{2}) & = (A \cap E_{1}) \cup (A \cap E_{2}) \\ = (A \cap E_{1}) \cup (A \cap E_{2} \cap E_{1}^{c}) \end{aligned}

Taking outer measure we get

\begin{aligned} ⟹ m^{*} (A \cap (E_{1} \cup E_{2})) & \leq m^{*} (A \cap E_{1}) + m^{*} (A \cap E_{2} \cap E_{1}^{c}) \\ E_{1} measurable ⟹ & = m^{*} (A) - m^{*} (A \cap E_{1}^{c}) + m^{*} (A \cap E_{2} \cap E_{1}^{c}) \\ = m^{*} (A) - m^{*} (A \cap (E_{1} \cup E_{2})^{c}) \end{aligned}

And this gives us the desired result.

Proof

by induction.
$n = 1$ is trivial. Suppose the statement holds for $n = k$ . Then

⋃_{j = 1}^{k + 1} E_{j} = (⋃_{j = 1}^{k} E_{j}) \cup E_{k + 1}

And then since unions of measurable sets are measurable we get the desired result.

File	Last Modified
Functional Analysis Lecture 7	2025-07-08
unions of measurable sets are measurable	2025-07-01

File

Last Modified

Functional Analysis Lecture 7

2025-07-08

unions of measurable sets are measurable

2025-07-01

unions of measurable sets are measurable

Statements

Single Unions

Countable Unions

References

See Also

Mentions