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$\sup_{-\infty<t<x} f(t)$ means?

Does it mean the least upper bound of the set of $f(t)$

OR the least upper bound of $t$ which will then be applied to $f(t)$?

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up vote 4 down vote accepted

It means the least upper bound of the set $\{f(t) | t \in (-\infty,x) \}$.

To see the difference, consider $f(x) = \arctan (-x)$. $f$ is strictly monotone decreasing. Then you can see that $\sup \{t | t \in (-\infty,x) \} = x$, but $\sup \{f(t) | t \in (-\infty,x) \} = 1$.

This also illustrates that $\sup \{f(t) | t \in (-\infty,x) \} \neq f(\sup \{t | t \in (-\infty,x) \})$.

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If it is of the set why it equals to $f(x-)$? – RHS Sep 23 '12 at 14:15
It is the least upper bound of the set, not the set. Also, it need not be $f(x-)$. – copper.hat Sep 23 '12 at 14:17
It is on p2 of "A Course in Probability Theory" is the word "monotonicity" mean anything here? – RHS Sep 23 '12 at 14:20
Yes, monotonicity is very relevant here. Monotone increasing means that if $t\leq t'$, then $f(t) \leq f(t')$. Similarly for decreasing, and if the $\leq$ is replaced by $<$, we have strict monotone increasing/decreasing. In this case, if $f$ is strictly monotone increasing, you will have $\sup \{f(t) | t \in (-\infty,x) \} = \lim_{t \to x} f(t)$. – copper.hat Sep 23 '12 at 14:23
So when will it equal to $f(x-)$? – RHS Sep 23 '12 at 14:25

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