Is this statement true or false? If true, how can I justify it with a proof. If false, is there a counter example. For any increasing functions $f : [a,b] \rightarrow R$, the set of discontinuities has measure zero.
For any increasing functions $f : [a,b] \rightarrow R$, the set of discontinuities has measure zero.
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There is a stronger statement: the set of discontinuities is countable. This is because any "uncountable sum" (see below) must diverge. Suppose otherwise. Then there is an increasing, uncountable sequence $x_I\in[a,b]$ of discontinuities of $f$. Here $I$ denotes an uncountable index set. What you now need to show is that $f$ necessarily blows up on $[a,b]$. So consider $f(b)$. You have that $f(b)\geq f(x)$ for all $x\leq b$. But:
$f(b)\geq f(a)+\sup_{C\in I: C\mbox{ is countable}} \sum_{i\in C}^\infty f(x_i)$.
Consider the set $S_n:=\{x_i: f(x_i)>=1/n\}$. Notice that $\cup_{n=1}^\infty S_n=\{f(x_i)\}_{i\in I}$. This implies that there exists an $N$ such that $S_N$ has infinitely many elements, so that the above supremum necessarily blows up.
Alex R.
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This shows that the set of discontinuities is countable. How can we ensure that the set of discontinuities has measure zero? – SRC94 Apr 08 '17 at 18:45
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Any countable set has measure 0: https://proofwiki.org/wiki/Countable_Sets_Have_Measure_Zero – Alex R. Apr 08 '17 at 18:46