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I was trying to prove this theorem-

"Every Absolutely continuous function is of bounded variation and hence is differentiable almost everywhere."

I got the first part like how Absolutely continuous function is of bounded variation but I am struggling to prove that if it of bounded variation then it is differentiable almost everywhere.

I tried to connect the fact that

$1)$ A function of bounded variation can be written as the difference of two increasing functions.

$2)$ A function $f:[a,b] \rightarrow \mathbb{R}$ be a monotone function then $f$ is differentiable almost everywhere.

But it is not working as difference of two increasing functions need not be a monotone function.

So how do I prove that a function of bounded variation implies it is differentiable almost everywhere?

Am I missing something?

Any help is great.

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As you have stated an AC function is of BV. A function of BV has a Jordan Decomposition. You can apply Lebesgues Differentiation Theorem to each function. The derivative of the sum is the sum of the derivative (Differential operator is linear). Both functions are differentiable almost everywhere, their sum is differentiable everywhere except the union of the two sets where each function is not differentiable. The union of two measure zero sets has measure zero.

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  • $\begingroup$ Nice one,now I know that we have to apply Lebegue theorem to each of the monotone increasing functions in the difference. $\endgroup$ – BAYMAX Apr 22 '17 at 4:57

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