Is $\inf_{[0,\:t]}f=\inf_{[0,\:t]\:\cap\:\mathbb Q}f$ even when $f$ is only left-continuous?

Question

Let $f:[0,\infty)\to\mathbb R$. If $f$ is continuous, then $$\inf_{[0,\:t]}f=\inf_{[0,\:t]\:\cap\:\mathbb Q}f\;\;\;\text{for all }t\ge0\tag1$$ by the intermediate value theorem. Can $(1)$ be shown in the case of left-continuity of $f$ too?

Answer 1

Yes. Suppose $L=\inf_{[0,t]} f$. If $f(0) = L$, we're done, since $0$ is rational.

Otherwise, for any $\epsilon>0$, there is some $c$ such that $c \in (0,t]$ and $|f(c)-L| < \frac{\epsilon}{2}$. In addition, by left-continuity, you can find some rational $d \in [0,c)$ such that $|f(d) - f(c)| < \frac{\epsilon}{2}$. Hence

$$\epsilon = \frac{\epsilon}{2} + \frac{\epsilon}{2}>|f(c)-L| + |f(d) - f(c)| \geq |f(d)-L|$$

In other words, there's rationals in $[0,t]$ which $f$ maps arbitrarily close to $L$.