From 16b67fe1e50fcc4484fc1de8b9362849f86a37f4 Mon Sep 17 00:00:00 2001 From: Anthony Berg Date: Thu, 13 Mar 2025 14:06:01 +0100 Subject: [PATCH] Quartz sync: Mar 13, 2025, 2:06 PM --- content/Lectures/Lecture 17 - Lp Spaces.md | 5 +---- 1 file changed, 1 insertion(+), 4 deletions(-) diff --git a/content/Lectures/Lecture 17 - Lp Spaces.md b/content/Lectures/Lecture 17 - Lp Spaces.md index db3a8170..db9c1893 100644 --- a/content/Lectures/Lecture 17 - Lp Spaces.md +++ b/content/Lectures/Lecture 17 - Lp Spaces.md @@ -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. For any finite subset $F$ of $X$, define -$$S_{F}(x) = \begin{cases} -g(x), & x\in F \\ -0, & x \not\in F -\end{cases}$$ +$$S_{F}(x) = \begin{cases}g(x), & x\in F \\ 0, & x \not\in F\end{cases}$$ Then $S_{F}$ is a [[Simple Function|simple function]], and $0\leq S_{F} \leq g$. Have $\int S_{F} \, d\mu = \int_{F} g \, d\mu = \sum_{x \in F} g(x)$ $S_{F} = \sum_{x \in F} g(x) X_{\{ x \}}$ since $\int S_{F} \, d\mu = \sum_{x \in F} g(x) \mu \underbrace{(\{ x \})}_{1}$