Let f,g : [a,b] $ \mathrm{\rightarrow} $ $ \mathrm{R} $ bounded and f $ \mathrm{\in}{\mathcal{R}}{\mathrm{[}}{a}{\mathrm{,}}{b}{\mathrm{]}} $ and $ \mathrm{D} $ $ \mathrm{{=}}{\mathrm{\{}}{\mathrm{x}}\mathrm{\in}{\mathrm{[}}{\mathrm{a}}{\mathrm{,}}{\mathrm{b}}{\mathrm{]}}\mathrm{\mid}{\mathrm{f}}{\mathrm{(}}{\mathrm{x}}{\mathrm{)}}\mathrm{\ne}{\mathrm{g}}{\mathrm{(}}{\mathrm{x}}{\mathrm{)}}{\mathrm{\}}} $ is of measure zero. Does it imply g $ \mathrm{\in}{\mathcal{R}}{\mathrm{[}}{a}{\mathrm{,}}{b}{\mathrm{]}} $?
My answer to it was;
Consider $h:[a,b] \mathrm{\rightarrow}{R}$ where $h(x)= f(x)-g(x)$ then the set of discontinuities of $h$ is $\mathrm{D}$, which is of measure zero; therefore by Lebesgue's criterion for Riemann integrability h$ \mathrm{\in}{\mathcal{R}}{\mathrm{[}}{a}{\mathrm{,}}{b}{\mathrm{]}} $.So, as h,f $ \mathrm{\in}{\mathcal{R}}{\mathrm{[}}{a}{\mathrm{,}}{b}{\mathrm{]}} $ so f-h = g $ \mathrm{\in}{\mathcal{R}}{\mathrm{[}}{a}{\mathrm{,}}{b}{\mathrm{]}} $.
Where did I go wrong in my argument, I don't understand. I do know that the answer to this question is that g$ \mathrm{\notin}{\mathcal{R}}{\mathrm{[}}{a}{\mathrm{,}}{b}{\mathrm{]}} $, I do have examples supporting the non integrability of g; but I just don't see where I went wrong. Any explanation or suggestion or solution is appreciated.