Monday, 17 February 2014

real analysis - Example $f$ Riemann-integrable, $g$ bounded and $g=f$ almost everywhere.

I'm facing this problem,





Let $g:[a,b]\rightarrow \mathbb{R}$ be Riemann-integrable, $f:[a,b]\rightarrow \mathbb{R}$ a bounded function, $(x_n)$ a sequence of points in $[a,b]$ such that $f(x)=g(x)$ for all $x$ in $[a,b]$ other than the $x_n$. $\textbf{Give an example}$ to show that $f$ need not be Riemann-integrable.




Before this, the book ("A First Course in Real Analysis by Sterling Berberian",Page 164) says that for $f:[a,b]\rightarrow \mathbb{R}$ and $g:[a,b]\rightarrow \mathbb{R}$ Riemann integrable, and $f=g$ almost everywhere we can conclude that $f$ is Riemann-integrable.



Checking the examples they give on non riemann integrability, I found $f(x):[0,1]\rightarrow \mathbb{R}$, $f(x)=1$ for rationals and $f(x)=0$ for irrationals. From this I can build a function $g:[0,1]\rightarrow \mathbb{R}$, that can be equal to $f$ over the irrationals, and something different than $f$ for rationals, However, how could I build a sequence such as the one they ask me to, but over the rationals?

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