Question

\mu(E_n) = \mu\{|f|/n \ge 1\} = \int I_{\{|f|/n \ge 1\}} d\mu \le \int |f|/n d\mu = 1/n \int |f| d\mu \xrightarrow{n \to \infty} 0 \\ where we used the trivial inequality \\ I_{\{|f|/n \ge 1\}} \le |f|/n

          \mu(E_n) = \mu\{|f|/n \ge 1\} = \int I_{\{|f|/n \ge 1\}} d\mu \le \int |f|/n d\mu = 1/n \int |f| d\mu \xrightarrow{n \to \infty} 0 \\ where we used the trivial inequality \\ I_{\{|f|/n \ge 1\}} \le |f|/n

μ(En) = μ{|f|/n ≥1} = ∫I{|f|/n ≥1} dμ≤∫|f|/n dμ= 1/n ∫|f| dμn →∞ 0
where we used the trivial inequality
I{|f|/n ≥1} ≤|f|/n

Added by Alba C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Vicki Stebbins

06/09/2020

Step 1

An indicator function is a function that takes the value 1 if a certain condition is true, and 0 if the condition is false. Show more…

Show all steps

Thanks for your feedback!

Hello, I'm reading a textbook on measure and I'm confused by the notation. Can someone explain what this I function is? (En)={lf|/nZI}= IfP/n21}d</1fl/ndu=1/n/1f|d"= where we used the trivial inequality I{Ifl/n>1} <|fl/n