Let $\{c_n\}$ a convergent sequence in $[a,b]$ and $f$ a function such that $f(x)=1$ if $x$ in $\{c_n\}$ and $0$ elsewhere. Prove $f$ is integrable. I'm working on a problem that goes as follows:

Let $a,b\in\mathbb{R}$ with $a<b$ and $\{c_n\}$ a convergent sequence in $[a,b]$. Let $f:[a,b]\rightarrow\mathbb{R}$ a function such that $f(c_n)=1$ for $n\geq0$ and $f(x)=0$ for all other $x\in[a,b]$.
a) Prove that for every $\epsilon>0$ there exists a partition $V$ with $U(V,f,x)-L(V,f,x)<\epsilon$.
b) Show that $f$ is riemann integrable.

Now b) isn't that hard, just play around with supremum and infinum. But for a) I built a proof that feels kinda convoluted, so I'd like it if someone could verify my proof. Both its correctness as well as its rigour.
Proof: Let $V$ be any partition $\{x_1, x_2, ..., x_n\}$ such that $x_1\neq x_2\neq ... \neq x_n$. Then by the denseness of $\mathbb{Q}$ in $\mathbb{R}$, $L(V,f,x) = 0$.
Let $\{c_n\}$ converge to $p\in[a,b]$. Then for any $\delta > 0$ there exists an integer $N$ such that $d(c_n,p)<\delta$ for $n>N$.
Consider a partition with $x_i=\max(p-\delta, a),\quad x_{i+1}=\min(p+\delta,b)$. Then - since $\delta$ is arbitrary - we can find a partition for every $\epsilon>0$ such that $\sup f(x) \Delta x_i<\epsilon$. This leaves us with finitely many intervals where $\sup f(x) \Delta x_k>0$. Choose some $\delta$ neighbourhood around each of these intervals such that $0<\delta<\frac{\epsilon}{N+1}$. Then $U(V,f,x)=\sum \sup f(x) \Delta x_k<\epsilon$ and thus, $U(V,f,x)-L(V,f,x)<\epsilon$.
In essence, I'm throwing the infinite terms of the convergence away by constructing some interval, and then using the usual method to let the sum approach $0$. I'm not entirely sure how clear this is though, so any tips are welcome.
 A: Just let $I=[p-\frac\epsilon4,p+\frac\epsilon 4]\cap[a,b]$. As you said, there are only finitely many, $N>0$ say, terms $c_i\notin I$. Around each of these, pick an interval of length $<\frac \epsilon{2N}$. Let $V$ be the partition obtained from the endpoints of these $N+1$ intervals. Then the total length of intervals where $f$ differs from $0$ is $<\frac \epsilon 2+N\frac{\epsilon}{2N}=\epsilon$, and the result follows.

Actually, to line up part a) suitably for a proof of b), given any $\epsilon>0$, note that there are only finitely many (again $N$, say) $i$ with $|c_i-p|>\frac\epsilon4$. Then for all $\delta$ with $\frac \epsilon 2+2\delta+N\delta<\epsilon$, it follows that for every partition $V$ finer than $\delta$, we have $U(V,f,x)-L(V,f,x)<\epsilon$:
Any difference occurs at most in the partition subintervals containing one of the $N$ outliers, and in subintervals partially overlapping $[p-\frac\epsilon4,p+\frac\epsilon 4]$. The former contribute at most $N\delta$, the latter at most $\frac\epsilon2+2\delta$ to $U(V,f,x)-L(V,f,x)$
A: ''Then by the denseness of $\mathbb{Q}$ in $\mathbb{R}$, $L(V, f, x) = 0$.''
I don't follow this particular reasoning, but if I understand what you're going for, there as a few ways to get there. If you know the Baire Category Theorem, you can say that the complement of a coutable set is dense basically immediately. If you don't know BCT, you can also do a slightly longer argument appealing to the fact that $(c_n)$ is a convergent sequence, thus it must have only one accumulation point, so all but at most one point (it's limit, if it occurs in the sequence) must be isolated, and so the complement of $\lbrace c_n \rbrace$ is dense.
