Continuity of Integral with Fixed Variables in Lebesgue Integration

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SUMMARY

The discussion focuses on the continuity of the function g(x) defined as g(x) = ∫01 f(x,y) dy, where f(x,y) is a continuous function of y for fixed x and vice versa, constrained by |f(x,y)| ≤ 1 for 0 ≤ x, y ≤ 1. The conclusion drawn is that g(x) is indeed continuous due to the properties of Lebesgue integration, which allows for the interchange of limits and integrals under certain conditions. The participant expresses confidence in their reasoning, emphasizing the continuity of f(x,y) and the bounded nature of the function.

PREREQUISITES
  • Understanding of Lebesgue integration
  • Familiarity with the concepts of continuity in multivariable functions
  • Knowledge of the epsilon-delta definition of continuity
  • Basic proficiency in mathematical analysis, particularly Rudin's Principles of Mathematical Analysis
NEXT STEPS
  • Study the properties of Lebesgue integrals and their implications for continuity
  • Explore the epsilon-delta definition of continuity in detail
  • Review examples of continuous functions in multiple variables
  • Examine theorems related to the interchange of limits and integrals in Lebesgue integration
USEFUL FOR

Mathematics students, particularly those studying real analysis, educators teaching Lebesgue integration, and researchers interested in the continuity of functions in multivariable calculus.

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Homework Statement



For reference, this is chapter 11, problem 12 of Rudin's Principals of Mathematical Analysis.

Suppose [tex]|f(x,y)| \leq 1[/tex] if [tex]0 \leq x \leq 1, 0 \leq y \leq 1[/tex]; for fixed x, f(x,y) is a continuous function of y; for fixed y, f(x,y) is a continuous function of x.

Put [tex]g(x) = \int_{0}^1 {f(x,y) dy}, 0 \leq x \leq 1[/tex].

Is g continuous?

Homework Equations



N/A

The Attempt at a Solution



To me it seems like this is obviously continuous since within an integral you're essentially working with a fixed y, so you can find some delta such that |f(x,y) - f(a,y)| is less than any positive epsilon, then the inequality you actually need follows easily. But it just seems way too easy.

Anyway, hope I didn't mangle the TeX. I'm not used to using it.

Just as a note, I'm referring to Lebesgue integration.
 
Last edited:
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But that epsilon depends on y.
 

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