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Math people:

My question is similar to that in Two definitions of "Bounded Variation Function" . If you look at that question, you will notice that people treat definitions (1) and (2) as the same, although (1) has the drawback that you can change $f$ on a set of measure zero and $V_a^b(f)$ changes. The answer posted to that question states the relationship between the two definitions. Is there any way to adjust definition (1) so it is equivalent to definition (2)? For example, $f$ is approximately continuous a.e. (see https://www.encyclopediaofmath.org/index.php/Approximate_continuity). What if you restricted the $x_j$'s to be points of approximate continuity?

Stefan (STack Exchange FAN)

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Stefan Smith
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You can normalize BV functions to be right-continuous. In this way, they behave as signed analogues to distribution functions of probability measure.

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  • could you be more specific or give a reference to how this is done? For example, how would you do it if the domain is $[0,]$, $f = 1$ on rational numbers and $f = 0$ for irrational numbers? – Stefan Smith Mar 12 '13 at 15:31
  • I think what he means it that you can amend the definition of $BV$ by requiring the function to be right-continuous. – Thomas Apr 15 '13 at 09:30
  • @Thomas : I posed a very similar question at http://math.stackexchange.com/questions/359946/can-i-make-a-bv-function-right-continous-this-way and eventually answered it myself. – Stefan Smith Apr 23 '13 at 23:00