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According to the intermediate value theorem, if $f$ is a real-valued continuous function on the interval $[a, b]$, and $u$ is a number between $f(a)$ and $f(b)$, then there is a $c ∈ [a, b]$ such that $f(c) = u$.

Now consider the function:

$$f(x) = x^3 - \lceil x \rceil$$

This function is obviously not continuous - it jumps whenever $x$ is integer. However, the jumps are always downward. Therefore, if $a<b$ and $f(a)<u<f(b)$, then there is a $c ∈ [a, b]$ such that $f(c) = u$ . I don't know how to prove it formally, but intuitively, the function cannot jump upwards, so in order to go from $f(a)$ up to $f(b)$, it must pass through all points in between.

My question is: Is there a theorem like this? If so, what is its name and what reference can I use for it?

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You may have luck searching for this using the conventional term "semi-continuous". –  Anthony Carapetis Aug 2 '13 at 9:20
    
Thanks: en.wikipedia.org/wiki/Semi-continuity So, my function is lower semi-continuous? –  Erel Segal Halevi Aug 2 '13 at 13:48
1  
It's upper semi-continuous. After a little thought it becomes apparent that it is not alone sufficient for the theorem you want; upper semi-continuous functions can jump in either direction depending on whether the function value at the jump is the left or right limit. Perhaps a combination of upper semi-continuous and "continuous from the left" would suffice. –  Anthony Carapetis Aug 2 '13 at 14:44
    
It's certainly true for piecewise continuous functions with all jumps being downwards - it's simple to show that one of the intervals of continuity $(p,q)$ has $f^+(p)<u<f^-(q)$ and then just apply IVT on this interval. –  Anthony Carapetis Aug 2 '13 at 14:49
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