Is the Product of Integrable Functions also Integrable?

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SUMMARY

The discussion centers on the integrability of the product of two functions, specifically $f(x,y)=h(x)g(y)$, where $h$ and $g$ are integrable functions on sets $X$ and $Y$. It establishes that if $f$ is integrable with respect to the product measure, then the integral of $f$ over the product space equals the product of the integrals of $h$ and $g$. This result holds without the need for $\mu$ and $\nu$ to be $\sigma$-finite, emphasizing the robustness of the theorem in measure theory.

PREREQUISITES
  • Understanding of integrable functions and their properties
  • Familiarity with product measures in measure theory
  • Knowledge of the concept of $\sigma$-finiteness in measures
  • Basic proficiency in mathematical notation and integration techniques
NEXT STEPS
  • Study the properties of product measures in detail
  • Explore the concept of integrability in the context of Lebesgue integration
  • Learn about the Fubini-Tonelli theorem and its applications
  • Investigate examples of integrable functions and their products
USEFUL FOR

Mathematicians, students studying measure theory, and anyone interested in advanced integration techniques will benefit from this discussion.

Chris L T521
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Here's this week's problem!

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Problem
: Let $h$ and $g$ be integrable functions on $X$ and $Y$, and define $f(x,y)=h(x)g(y)$. Show that if $f$ is integrable on $X\times Y$ with respect to the product measure, then
\[\int_{X\times Y} f\,d(\mu\times\nu) = \int_X h\,d\mu \int_Y g \,d\nu.\]

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Note: We do not need to assume that $\mu$ and $\nu$ are $\sigma$-finite.

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered this week's problem. I don't have a solution ready at this time; it's on my backlog of solutions to write and I'll try to get them done soon. I take the GRE on Saturday so it's been crunch time for me. Thank you for your patience!
 

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