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Anthony Berg 2025-03-13 14:06:01 +01:00
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content/Lectures/Lecture 17 - Lp Spaces.md
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@ -101,10 +101,7 @@ Say $g : X \to [0, \infty \rangle$.
Then (define: $s = \sum_{i = 1}^{n} a_{i} X_{A_{i}}$) $\int g \, d\mu = \sup_{0 \leq s \leq g} \int s \, d\mu = \sup_{0\leq s\leq g} \sum_{i=1}^{n} \overbrace{a_{i}}^{\neq 0} \times \mu(A_{i}) = \infty$ if any $A_{i}$ is infinite. Then (define: $s = \sum_{i = 1}^{n} a_{i} X_{A_{i}}$) $\int g \, d\mu = \sup_{0 \leq s \leq g} \int s \, d\mu = \sup_{0\leq s\leq g} \sum_{i=1}^{n} \overbrace{a_{i}}^{\neq 0} \times \mu(A_{i}) = \infty$ if any $A_{i}$ is infinite.
For any finite subset $F$ of $X$, define For any finite subset $F$ of $X$, define
$$S_{F}(x) = \begin{cases} $$S_{F}(x) = \begin{cases}g(x), & x\in F \\ 0, & x \not\in F\end{cases}$$
g(x), & x\in F \\
0, & x \not\in F
\end{cases}$$
Then $S_{F}$ is a [[Simple Function|simple function]], and $0\leq S_{F} \leq g$. Have $\int S_{F} \, d\mu = \int_{F} g \, d\mu = \sum_{x \in F} g(x)$ Then $S_{F}$ is a [[Simple Function|simple function]], and $0\leq S_{F} \leq g$. Have $\int S_{F} \, d\mu = \int_{F} g \, d\mu = \sum_{x \in F} g(x)$
$S_{F} = \sum_{x \in F} g(x) X_{\{ x \}}$ since $\int S_{F} \, d\mu = \sum_{x \in F} g(x) \mu \underbrace{(\{ x \})}_{1}$ $S_{F} = \sum_{x \in F} g(x) X_{\{ x \}}$ since $\int S_{F} \, d\mu = \sum_{x \in F} g(x) \mu \underbrace{(\{ x \})}_{1}$