Let $f,g\in \scr{R}[a,b]$ such that measure of $A=\{x\in [a,b]|f(x)\neq g(x)\}$ is zero. Show that $$\int_{a}^{b}f(x)\,\mathrm{d}x=\int_{a}^{b}g(x)\,\mathrm{d}x$$
Now, application of machineries from measure theory is prohibited. So, I was thinking in the lines of this : There are intervals $\{J_s\}_{s\in \mathbb{N}}$ such that $\displaystyle A\subseteq \bigcup_{s\in \mathbb{N}}J_s$ and such that $\displaystyle \sum_{s\in \mathbb{N}}\ell(J_s)<\varepsilon$ here $\ell(\cdot )$ denotes length of an interval. Now, I need a partition from it. But, if I had a finite subcover, then we would have generated a partition but this is not helping. And even if we had a finite subcover, I intuitively feel there is a partition but it all breaks down when I try to write it down. So, I am not sure about my ways. Can someone drop some hints? Thanks a lot.
This is adapted from theorem 8.3.4 in Elementary Classical Analysis by Marsden & Hoffman.
Let $h=f-g$. Then $h$ is equal to $0$ except on the set $A$. $h$ is Riemann integrable, and in particular bounded. Let $M$ be such that $|h(x)|\le M$ for all $x\in[a,b]$. Let $P$ be a partition of $[a,b]$ in subintervals $I_1,\dots,I_n$ and let $$ L(h,P)=\sum_{i=1}^Nm_i(h)\,|I_i| $$ be the lower sum, where $m_i(h)=\inf_{x\in I_i}h(x)$ and $|I_i|$ is the length of $I_i$. Then $m_i(h)\le M\,m_i(\chi_A)$,where $\chi_A$ is the characteristic function of $A$ and $$ L(h,P)\le M\sum_{i=1}^Nm_i(\chi_A)\,|I_i|. $$ The only way in which $m_i(\chi_A)$ can be different from $0$ is if $I_i\subset A$. But this is impossible, since a set of measure $0$ cannot contain an interval (this is a non trivial fact). Thus $$ L(f,P)\le0\implies\int_a^bh\le0. $$ We can prove similarly that the upper sums are non-negative, and that $$ \int_a^bh\ge0. $$