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Last edited by hbghlyj 2022-10-18 13:47Showing that the "left-hand limit function" is left continuous
Use the sequential definition (not the $\epsilon-\delta$ definition) of left continuity to prove this question.
In other words, let $(s_n)$ be a sequence contained in $(0, \infty)$ such that $s_n<t$ for all $n$ and $s_n \rightarrow t$. I need to show that $\lim_{n \rightarrow \infty} f^-(s_n) = f^-(t) = f(t^-)$. I have also been given a hint (but I have no idea where to incorporate it):
If $f(t^-)$ exists and is finite, then it is equivalent to:
$$f(t^-) = \sup \inf_{s<t} \{f(v): s \le v < t\} = \inf \sup_{s <t} \{f(v) : s \le v < t\} $$
Using the definition of $f^-(t)$, given $\varepsilon>0$ there is $\delta>0$ such that $$|f(s)-f^-(t)|\le\varepsilon $$
for all $t-\delta\le s<t$. Since $s_n\to t^-$, there is $\bar n$ such that $t-\delta<s_n<t$ for all $n\ge \bar n$.
Fix $n\ge \bar n$. Then for $t-\delta<s<s_n$ you have $$|f(s)-f^-(t)|\le\varepsilon $$ or equivalently,
$$f^-(t)-\varepsilon\le f(s)\le f^-(t)+\varepsilon. $$
Letting $s\to s_n^-$ in the previous inequality, you get
$$f^-(t)-\varepsilon\le f^-(s_n)\le f^-(t)+\varepsilon. $$
So you have $|f^-(s_n)- f^-(t)|\le \varepsilon$ for all $n\ge \bar n$. |
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