If $x:[0,\infty)$ is càdlàg, then the left-limit function $x^-(t):=\lim_{s\to t-}x(s)$ has right-limits equal to $x(t)$

Solution 1:

In fact, the proof is elementary and just plays with definition.

Let $t_{0}\geq0$ be fixed. Denote $x(t_{0})=L$. Let $\varepsilon>0$ be arbitrary. Choose $\delta>0$ such that $d(x(t),L)<\varepsilon$ whenever $t\in[t_{0},t_{0}+\delta)$. This is possible because $x$ is right-continuous.

Let $t\in(t_{0},t_{0}+\delta)$ be arbitrary. By the definition of left-limit $x^{-}$, there exists $s\in(t_{0},t)$ such that $d\left(x^{-}(t),x(s)\right)<\varepsilon$. It follows that \begin{eqnarray*} & & d\left(x^{-}(t),L\right)\\ & \leq & d\left(x^{-}(t),x(s)\right)+d\left(x(s),L\right)\\ & < & 2\varepsilon \end{eqnarray*} by observing that $s\in(t_{0},t)\subseteq[t_{0},t_{0}+\delta)$. Therefore $\lim_{t\rightarrow t_{0}+}x^{-}(t)=x(t_{0})$.