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$f\in\mathcal{R}[a,b], f=g$ almost everywhere $\implies$ $g\in\mathcal{R}[a,b]$?

Attempt/Thoughts:

$f=g$ almost everywhere so that means they are equal everywhere, except on the sets where $f$ has measure zero. So let's assume that on these intervals where $f$ has measure zero, $g$ actually has non-zero measure. I know that Lebesgue's criterion for Riemann Integration gives us that the measure of the discontinuities of a function have to be equal to $0$ in order for a function to be Riemann Integrable. So the real question is if the sets where we have measure zero are replaced with intervals where we don't have measure zero, is the measure of the discontinuities still zero? I think this forces $g\notin\mathcal[a,b]$ because it changes sets of measure zero, and Lebesgue's criterion requires the measure of the discontinuities to be zero, but we are altering this set to have non-zero measure.

Similarly, if instead of $f=g$ almost everywhere, we had that $f=g$ except at countably many points, could we use the fact that the measure of a countable set is equal to zero to draw conclusions about the measure of the discontinuities of $g$ to determine R. Integrability by Lebesgue's criterion?

Thanks.

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  • $\begingroup$ $f=g$ almost everywhere means that $f=g$ everywhere except on a set of measure $0$. "on the sets where f has measure zero" this sentence doesn't makes sens to me. What is the measure of a function? Its integral? $\endgroup$ – Surb Feb 2 '15 at 8:50
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    $\begingroup$ Note the set of rationals has measure zero. So the constant function $f=1$ is equal to the characteristic function of the irrationals a.e.. $\endgroup$ – David Mitra Feb 2 '15 at 8:52
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consider $f\equiv0$

$g(x)=1 on \mathbb Q$ and $0$ on $\mathbb Q^c$

Here $f$ is $R$ integrable but $g$ is not

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