Riemann integrable functions continuous except on a set of measure zero?

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SUMMARY

A function is Riemann integrable on a bounded interval if its set of discontinuities has measure zero, as established by the Riemann-Lebesgue lemma. However, it is incorrect to assert that such a function must be equal to a continuous function almost everywhere. The example provided illustrates that the function f(x) defined as 1 for 0 ≤ x ≤ 1/2 and 0 for 1/2 < x ≤ 1 is continuous almost everywhere but does not equal a continuous function almost everywhere. Nevertheless, for any ε > 0, there exists a set A of measure less than ε and a continuous function g such that f and g are equal outside of A.

PREREQUISITES
  • Understanding of Riemann integrability
  • Familiarity with the concept of measure zero
  • Knowledge of the Riemann-Lebesgue lemma
  • Basic principles of continuity in real analysis
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  • Explore examples of functions with discontinuities and their integrability
  • Learn about Lebesgue integration and its advantages over Riemann integration
  • Investigate the concept of sets of measure zero and their properties
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Students and professionals in mathematics, particularly those studying real analysis, as well as educators looking to clarify concepts related to Riemann integrability and continuity.

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Is it true that a function is Riemann integrable on a bounded interval only if it's equal to a continuous function almost everywhere? I'd imagine this is the case, given the Riemann-Lebesgue lemma, which says that a function is RI iff its set of discontinuities has measure zero. (So the "continuous function" is then just f restricted to the complement of its set of discontinuities.) But I might be wrong. Help?
 
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I've just discovered this is incorrect. Consider the function

<br /> f(x) = \begin{cases}<br /> 1 &amp; \text{ if } 0\leq x \leq 1/2\\<br /> 0 &amp; \text{ if } 1/2 &lt; x \leq 1<br /> \end{cases}<br />

Then f is continuous almost everywhere, but it cannot be equal to a continuous function almost everywhere by an argument involving inverse images of open sets, etc. Bummer.
 
However, for every e>0, there exists a set A of measure less than e and a continuous function g such that, outside of A, f and g are equal.
 

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