Let $u,v : [a,b] \to \mathbb R$ be continuous functions , let $f(x):=u(x)$ , when $x$ is rational ; $f(x):=v(x)$ , when $x$ is irrational . If $f$ is Riemann integrable , then is it true that $u(x)=v(x) , \forall x \in [a,b]$ ? If it is true , then I would like a proof possibly avoiding the Lebesgue Integrability criteria of measure $0$ . The only thing I have figured out is that $f$ is continuous at $x_0$ iff $u(x_0)=v(x_0)$ , then to verify the integrability ' with ' Lebesgue criteria , it appears , if $g\in C[a,b]$ is such that $g(x)=0$ a.e. then $g$ is identically $0$ ?
Please help . Thanks in advance
If $f$ is Riemann integrable on $[a,b]$, then it is Riemann integrable on each subinterval of $[a,b]$. Assume that you can find $x_0 \in [a,b]$ for which $u(x_0) \neq v(x_0)$. Without loss of generality, assume that $u(x_0) > v(x_0)$. Then, by continuity, you can find some $\varepsilon > 0$ such that $u(x) > v(x)$ for all $x \in [x_0 - \varepsilon, x_0 + \varepsilon]$. If you consider the Riemann sums of $f|_{[x_0 - \varepsilon, x_0 + \varepsilon]}$that correspond to partitions and choices of rational (irrational) points then, as the diameter of the partitions goes to zero, they will converge to $\int_{x_0 - \varepsilon}^{x_0 + \varepsilon} u(x) \, dx$ (or $\int_{x_0 - \varepsilon}^{x_0 + \varepsilon} v(x) \, dx $ respectively). Since $f$ is integrable by assumption, we must have
$$ \int_{x_0 - \varepsilon}^{x_0 + \varepsilon} u(x) \, dx = \int_{x_0 - \varepsilon}^{x_0 + \varepsilon} v(x) \, dx $$
contradicting the fact that $u(x) > v(x)$ on $[x_0 - \varepsilon, x_0 + \varepsilon]$.