I'm trying to find a sequence of continuous functions $f_n: [0,1] \rightarrow \mathbb{R}$ such that for every $x\in [0,1]$ the sequence $f_n(x)$ is monotonic and convergent, and that the limit function $f$ isn't Reimann-integrable on $[0,1]$.
My attempt was as follows:
First fix $0<\epsilon<1$. Then take an enumeration of $\mathbb{Q}\cap[0,1]$ and denote the $n$th rational as $q_n$.define $g_n$ to be equal $1$ everywhere except for a continuous 'dip' of width $\frac{\epsilon}{2^n}$ around $q_n$ such that $g_n(q_n)=0$. Now define $f_1 = g_1,f_n=min(g_n,f_{n-1})$.
It is obvious that the sequence $f_n$ is pointwise monotone and bounded below by $0$, hence also convergent. It remains to show the non-integrability of the limit.
Define $A_0 =f^{-1}(\{0\}), A_1=f^{-1}(\{1\})$. My sketch is:
$\mathbb{Q}\cap[0,1]\subseteq A_0$ so $A_0$ is dense. The total width of the 'dips' does not exceed $\epsilon<1$ so $A_1$ is of positive measure $\rightarrow ?$ $A_1$ isn't nowhere-dense $\rightarrow$ there is a subinterval $[a,b]$of $[0,1]$ such that $A_0$ and $A_1$ both intersect every subinterval of $[a,b]$ $\rightarrow$ $f$ isn't Riemann-integrable.
The part marked with $'?'$ is the one I'm unsure about. Am I on the right track?
Additionally, I am just starting my study of measure-theory, so any way to avoid using measure-theoretic arguments altogether is most welcome.