The Riemann Integrability of a function similar to Dirichlet's function

The Riemann Integrability of a function similar to Dirichlet's function

$\textbf{Problem:}$ Consider the function $f: [0,1] \rightarrow
\mathbb{R}$ defined by letting $f(x)=0$ for rational $x$ and $f(x)=x$ for
irrational $x$. Calculate the upper and lower Riemann integrals of $f$. Is
$f$ Riemann integrable?
$\textbf{ Solution Attempt:}$ Let $P_n=\{0=t_0,...,t_n=1\}$ be a partition
of $[0,1]$, where $t_k = \frac{k}{n}$. For this partition, we have
$$U(f,P_n)=\sum^{n}_{k=1} t_k (t_k - t_{k-1}) = \sum^{n}_{k=1} t_k
(\frac{k}{n} - \frac{k-1}{n})=\sum^{n}_{k=1} \frac{k}{n} (\frac{1}{n})=
\frac{1}{n^2} \sum^{n}_{k=1} k.$$ A familiar property of the natural
numbers gives us that $\sum^{n}_{k=1} k= \frac{1}{2}n(n+1).$ Thus,
$U(f,P)= \frac{1}{2} + \frac{1}{2n}.$ Recall, the upper Riemann integral
$U(f)$ is defined as $U(f)=\inf \{U(f,P)\}$, where $P$ is a parition of
$[0,1].$ As $n \rightarrow \infty$, $P_n \rightarrow \frac{1}{2}.$ [Note:
this is where I am a little unsure of myself.] Thus, $U(f)=\frac{1}{2}$.
However, the lower Riemann integral is $0$. Hence, the upper and lower
Riemann integrals disagree. We conclude that $f$ is not Riemann
integrable. $\blacksquare$
If the aforementioned function was changed so that it took on the value of
$0$ for irrational $x$ and $x$ for rational $x$, then $f$ would be zero
almost-everywhere. Hence, it would be Lebesgue integrable. Is this
correct? I am quite unsure of the Lebesgue integrability for the original
function. A hint would great! Thank you!