**Proposition 1** Suppose that $f(t):\mathbb{R}_+\mapsto\mathbb{R}$ is left continuous, then we can define a sequence of step functions $f_m(t)$ by $$f_m(t)=\sum_{k\geq 1}f_{m,k}(t),$$ where $$f_{m,k}(t) = f( (k-1)2^{-m} ),\hspace{10pt}\forall\, t\in ((k-1)2^{-m},k2^{-m} ].$$ Then $f_m$ converges to $f$ point-wise. **Remark 1** For a right continuous function, we can define $f_m$ in a similar fashion with $$f_{m,k}(t) = f( k2^{-m} ),\hspace{10pt}\forall\, t\in [(k-1)2^{-m},k2^{-m} ).$$ Then we can also show that $f_m$ converges to $f$ point-wise. //Proof of Proposition 1// Fix any $t_0\in \mathbb{R}_+$, by the left continuity of $f$, we know that for any $\ep>0$, there exists a $\delta>0$ such that $$(1)\hspace{30pt}\abs{f(t)-f(t_0)} < \ep,\hspace{10pt}\forall\, 0\leq t_0 - t<\delta.$$ Now choose a large $m_0$ such that $2^{-m_0}<\delta$ and assume that $t_0\in ((k_0-1)2^{-m_0},k_02^{-m_0} ]$, then by $(1)$, we have $$(2)\hspace{30pt}\abs{ f_{m_0}(t_0) - f(t_0) }=\abs{ f((k_0-1)2^{-m_0}) - f(t_0) } < \ep.$$ It is easy to see that $(2)$ also holds for any $m>m_0$. We thus completed the proof. $\square$