There are numerous functions which are Lebesgue but not Riemann integrable, the most famous one probably being

$$ f: [0,1] \rightarrow \mathbb{R}, \quad x \mapsto \begin{cases} 1, & x \in \mathbb{Q} \\ 0, & \text{otherwise} \end{cases}$$

However, changing $f$ on a null set (namely $\mathbb(Q) \cap [0,1]$) could result in a Riemann integrable function.

My question is: Is this true for all Lebesgue but not Riemann integrable functions? If not, are there any counter examples constructable?

  • $\begingroup$ $f(x)=1/\sqrt x$ will do (though perhaps isn't very interesting). $\endgroup$ Jan 16, 2014 at 18:03

1 Answer 1


Yes, the characteristic function of a fat cantor set is not almost equal to any Riemann integrable function.

  • $\begingroup$ After wikipedia–ing “fat cantor set” this sounds plausible, thanks. :) $\endgroup$
    – Keba
    Jan 16, 2014 at 23:35

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