Let $f$ be a continuous real-valued function on $(0,1]$. Show that \begin{equation} \int_{[0,1]} |f| \,dm = \lim_{n \rightarrow \infty} \int_{1/n}^1 |f(x)| \,dx \end{equation}
where the integral $\displaystyle{\int_{1/n}^1 |f(x)| \,dx}$ is the Riemann integral.
Here $m$ is the measure on $(0,1]$ satisfying $m(a,b) = b-a$.
My understanding of the formalities here is not very good. I am tempted to say that as $n \rightarrow \infty$ it follows that $1/n \rightarrow 0$ and so we have in the limit that:
\begin{equation} \lim_{n \rightarrow \infty} \int_{1/n}^1|f(x)|\,dx = \int_0^{\infty} |f(x)|\,dx = \int_{[0,1]}|f|\,dm \end{equation}
where the last equality follows as the Riemann integral exists, it must be equal to the Lebesgue integral.
I feel uneasy about this proof however as I have not used the hypothesis that $f$ is continuous on $(0,1]$. Can someone help with the formalities behind this proof?