If $f : [0, 1] → \mathbb R$ is a function such that $f(x) = 0$ for $x = 0$ and $f(x) = 2$ for $x ∈ (0, 1]$. Show that $f$ is Riemann integrable on $[0, 1]$.
Here, I thought to divide the partition in to equal subintervals and consider two cases, where $0 \in [x_{k-1} , x_{k}]$ and $0$ not in the $k^{th}$ subinterval.
when $x \neq 0 , f(x) = 2 $ hence $U(f) = L(f) = 2$ when $0 \in [x_{k-1} , x_{k}] , L(f) = 0$ and $U(f) = sup \{\dfrac{2}{n} \}$ the limit is still $0$
is there a better way to say this is Riemann Integrable?