Is there a sequence $(f_n)$ of Riemann integrable functions such that $\lim f_n(x) = f(x)$ almost everywhere on $[a,b]$ and $\lim\int_a^bf_n$ does not exists in Riemann sense, but it does in Lebesgue sense?
By the way the author said, is a simple example. But I waste a lot of time trying to find and nothing appear...
(please, apologize the bad english)
On
Take $(e_i)_{i \in \mathbb{N}}$ enumeration of the rationals in $[0,1]$.
Define a sequence $f_n: [0,1] \rightarrow \mathbb{R}$ as follows:
$f_n(x)=1$ if $x \in \{e_1,...,e_n \}$ and $f_n(x)=0$ otherwise.
Each $f_n$ is continuous except at a finite number of points, hence riemann integrable. But the pointwise limit function is not riemann integrable, being the indicator function of the rational numbers in $[0,1]$.
NOTE THAT: This isn't what you wrote in your question, although I think this is what you intended. $\lim \int f_n$ is the limit of a sequence of number, by your own hypothesis on the $f_n$. This answer assumes (as I think it is the case) that you want a sequence of riemann integrable functions for which the pointwise limit is not riemann integrable.
Each $\ln_a^b f_n$ is simply a number; so if each number is defined but the sequence of numbers doesn't converge, then extending from Riemann integrals to Lebesgue integrals (which will all have the same value) doesn't change anything. If you want an example where $\int_a^b (\lim f_n)$ doesn't exist as a Riemann integral but does exist as a Lebesgue integral, then let $\{q_j\}$ be an enumeration of the rational numbers in $(a,b)$ and let $f_n(x) = 0$ except that $f_n(q_1)=\cdots=f_n(q_n)=1$.