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Show that every function of bounded variation is discontinuous in at most a countable set of points.

This next proof was declared wrong by a teacher, but I am still not able to realize the wrong assumptions, and it only partially helps me to know the correct proof. Does anyone have an observation about what is wrong in the proof or "how bad" is the proof. THanks

Proof:

Let $g(x), h(x) : [a,b] \rightarrow \Bbb{R} $ be increasing monotonous functions.

Let $P_1$ be a partition of $[a,b]$ such that, if $\bar{x_i} \in P_1$, then in $g(\bar{x_i}) $ there exists a jump discontinuity.

Let $P_2$ be a partition of $[a,b]$ such that, if $\bar{x_i} \in P_2 $, then in $h(\bar{x_i}) $ there exists a jump discontinuity.

As $g,h$ are increasing monotonous functions, they are of bounded variation, so, as I defined them, $g(\bar{x_i}) $,$h(\bar{x_i}) $ are the only discontinuities for $g,h$.

As $P_1$ ,$P_2$ Are finite partitions, $g,h$ are continuous except for a countable set of points.

Let $$ P = \{ a=x_0, x_1, ..., x_n = b \} $$ be a partition of $[a,b]$ such that $P=P_1 U P_2$,

Be cause of a theorem, if $f:[a,b] \rightarrow \Bbb{R} $ is of bounded variation, then there exist funtions $g,h :[a,b] \rightarrow \Bbb{R} $ such that $f(x) = g(x) - h(x)$ where g,h are increasing monotonous functions.

I pick precisely $f,g,h$ as defined above.

As $g,h$ are continuous except for a countable set of points, $f $ will be continuous except for a countable set of points.

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    $\begingroup$ Of what use are those partitions? The only thing needed is $D(f)\subset D(g)\cup D(h)$ and since both $D(g),D(h)$ are countable, so is their union. (Here D denotes the set of discontinuities.) $\endgroup$ – zhw. Nov 16 '16 at 18:06

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