Let $f$ be continuous on $[0,1]$.
$$\lim_{n\rightarrow \infty} \sum_{j=0}^{[n/2]} \frac{1}{n} f\left(\frac{j}{n}\right)$$
where [n] is greatest integer function.
I used the following relation,
$$\int_{a}^{b}f(x)\ dx =\lim_{\delta_x \rightarrow 0} \sum_{x=a}^{b}f(x)\delta_x$$
I considered $x=\frac{j}{n}$ and $\delta_x = \frac{(j+1)-j}{n}=\frac{1}{n}$.
At $j=0$, I've $x=0$.
At $j=[n/2]$, if $n$ is even then $x=\frac{1}{2}$. If $n$ is odd,$[n/2]=\frac{n-1}{2}$ then $x=\frac{1}{2}[1-\frac{1}{n}]$.
Therefore, as $n\rightarrow \infty$ out $x=\frac{1}{2}$.
$$\lim_{n\rightarrow \infty} \sum_{j=0}^{[n/2]} \frac{1}{n} f\left(\frac{j}{n}\right)=\int_{0}^{1/2} f(x)\ dx$$