Proving Continuity of Compositions: Sets of Measure Zero

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SUMMARY

The discussion focuses on proving that if a function f is continuous almost everywhere (a.e.) on the interval [a,b] and a function g is continuous a.e. on [c,d], with the image of f contained in [c,d], then the composition g(f(x)) is also continuous a.e. The proof initially relied on the Riemann integrability of both functions, but the author recognized a potential flaw in assuming that the composition of Riemann integrable functions remains Riemann integrable. Ultimately, the conclusion is that the points of discontinuity of g(f(x)) are limited to the union of the discontinuities of f and g, which are both null sets, confirming that g(f(x)) is continuous a.e.

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  • Understanding of continuity and Riemann integrability
  • Knowledge of measure theory, specifically sets of measure zero
  • Familiarity with function composition in real analysis
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  • Study the properties of Riemann integrable functions
  • Explore measure theory, focusing on null sets and their implications
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Homework Statement



I am trying to prove that if f is continuous almost everywhere on [a,b], and if g is cont a.e. on [c,d], with
f[a,b] contained in [c,d], then g composite f is cont. a.e.


The Attempt at a Solution


------

Originally, my proof went something like this:

f is cont. a.e. implies f is Riemann integrable

g is cont. a.e. implies g is Riemann integrable

since f and g are Riemann integrable, g composite f is Riemann integrable (*)

A Riemann integrable function is cont. a.e., thus g composite f is cont. a.e.

-------

I was satisfied with this until I realized (*) might not be true, as I couldn't prove it as a Lemma.

Also, I am a newby to this forum - does anyone know if there is a way to LaTex my input?

Thanks
 
Physics news on Phys.org
The points of discontinuity of g composite f are contained in the union of those of f & g.
Since the union of two null sets is also null, gof will be continuous a.e.
 

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