# Difference of increasing functions differentiable a.e.

I'm working through Royden & Fitzpatrick's Real Analysis, in the beginning of section 6.3 it reads and I quote: "Lebesgue's theorem tells us that a monotone function on an open interval is differentiable almost everywhere. Therefore the difference of two increasing functions on an open interval also is differentiable almost everywhere."

For this to be true, it means that the difference of two increasing functions on an open interval is monotone, but I do not see how that is immediately obvious.

Any clarification is appreciated.

• No! It says the difference of two increasing functions is differentiable everywhere. It does not have to be monotone. – user99914 Nov 16 '15 at 7:55
• @JohnMa: Thx for your comment, then how does he arrive at that conclusion, and why does he place that statement right after restating the Lebesgue theorem? – Mike Nov 16 '15 at 7:56
• Because this is an immediate consequence of that theorem. If $f_1, f_2$ are differentiable almost everywhere, then so is $f_1 - f_2$. – user99914 Nov 16 '15 at 7:57
• @JohnMa: Hmm, I'm sorry I don't quite see that. The theorem states "$f$ monotone $=>$ $f$ differentiable a.e." From this information the only way I can claim $f=f_1 - f_2$ is differentiable a.e. is by showing $f$ is monotone. – Mike Nov 16 '15 at 8:02
• If $f_1$, $f_2$ are differentiable at $a$, then $f_1 - f_2$ is also differentiable at $a$ (nothing to do with measure theory). Also the set $\{a : f_1, f_2\text{ is differentiable at } a\}$ is of full measure. – user99914 Nov 16 '15 at 8:08