关于递增右连续函数左逆的下确界等式$A_{s-}^{-1}=\inf\{t\geq 0: A_t\geq s\}$的证明求助

Hey there! Let's work through this proof step by step to close the gap on that missing inequality direction.

First, let's recap the definitions to keep us aligned:

• $s \mapsto A_s$ is increasing and right-continuous for $s \geq 0$.
• $A_s^{-1} := \inf{t \geq 0 : A_t > s}$ (with $\inf{\emptyset} = \infty$), which is also increasing.
• $A_{s-}^{-1} := \lim_{h \to 0^+} A_{s-h}^{-1}$ (the left limit of the inverse function at $s$).

You already showed that $\lim_{h \to 0^+} A_{s-h}^{-1} \leq \inf{t \geq 0 : A_t \geq s}$—great start! Now let's prove the reverse inequality: $\inf{t \geq 0 : A_t \geq s} \leq \lim_{h \to 0^+} A_{s-h}^{-1}$.

Step 1: Notation and key properties #

Let $T = \inf{t \geq 0 : A_t \geq s}$ and $L = \lim_{h \to 0^+} A_{s-h}^{-1}$. Since $A_s^{-1}$ is increasing, $A_{s-h}^{-1}$ is increasing as $h \to 0^+$ (because $s-h$ increases as $h$ shrinks, and the inverse of an increasing function is increasing). This means $L$ is the supremum of ${A_{s-h}^{-1} : h > 0}$, so $A_{s-h}^{-1} \leq L$ for all $h > 0$, and we can make $A_{s-h}^{-1}$ arbitrarily close to $L$ by choosing small enough $h$.

Step 2: Assume the opposite (for contradiction) #

Suppose for contradiction that $T > L$. Then there exists some $t_0$ with $L < t_0 < T$. By the definition of $T$ (the infimum of all $t$ where $A_t \geq s$), for all $t \leq t_0$, we have $A_t < s$.

Step 3: Use right-continuity to derive a contradiction #

Since $A$ is right-continuous at $t_0$, $A_{t_0} = \lim_{t \to t_0^+} A_t \leq s$. Let $h_0 = s - A_{t_0} > 0$ (this is positive because $A_{t_0} < s$). Then $s - h_0 = A_{t_0}$.

Because $A$ is increasing, for all $t \leq t_0$, $A_t \leq A_{t_0} = s - h_0$. This means the set ${t \geq 0 : A_t > s - h_0}$ can only contain $t > t_0$. Therefore:
$$A_{s-h_0}^{-1} = \inf{t \geq 0 : A_t > s - h_0} \geq t_0$$

But $t_0 > L$, and we know $A_{s-h_0}^{-1} \leq L$ (since $L$ is the supremum of all $A_{s-h}^{-1}$). This is a contradiction!

Step 4: Combine both inequalities #

Our contradiction shows that $T \leq L$, i.e., $\inf{t \geq 0 : A_t \geq s} \leq \lim_{h \to 0^+} A_{s-h}^{-1}$.

Combined with your initial result:
$$\lim_{h \to 0^+} A_{s-h}^{-1} \leq \inf{t \geq 0 : A_t \geq s} \leq \lim_{h \to 0^+} A_{s-h}^{-1}$$

This forces the equality:
$$A_{s-}^{-1} = \inf{t \geq 0 : A_t \geq s}$$

备注：内容来源于stack exchange，提问作者user123234