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Let f be a monotone increasing function on [0,1] with f(0)=0 and f(1)=1. Let E = f([0,1]) and the Lebesgue measure mE = 1. Prove f is continuous on [0,1].

I've been attempting this problem at several angles, but cannot seem to work with it. Is it wise to work with discontinuities and their measure?

Brady
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    Monotone functions are continuous almost everywhere, so the set of discontinuities has measure $0$, now you just need to show it is actually empty. – Dylan Yott Dec 09 '13 at 03:58
  • @DylanYott To say that the set of discontinuities of a monotone function has measure $0$ is an understatement; the set of discontinuities is countable. I don't see, though, how that fact is relevant to this question. – bof Dec 09 '13 at 04:30
  • @bof, good point, I tried to solve this using that fact and it didn't go anywhere, so I left it as a comment hoping it would be useful to someone. – Dylan Yott Dec 09 '13 at 04:43

2 Answers2

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Any discontinuity of a monotone function is a jump discontinuity, meaning that $f(a-)\lt f(a)$ or $f(a)\lt f(a+)$. Either way, a whole interval is missing from the range of $f$, and so $m(E)$ falls short of $1$ by at least the length of that interval.

bof
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The only type of discontinuity a monotone function can have are jumps. First note that $\displaystyle f[0,1]\subset [0,1]$. Now pick a point $y\in [0,1]$. Then $\displaystyle 1=m(f[0,1])\le 1-\lim_{x\to y^+}|f(x)-f(y)|\le 1$, so that $\lim_{x\to y+}|f(x)-f(y)|=0$, and similarly $\lim_{x\to y^-}|f(x)-f(y)|=0$, so that $f$ is continuous.

JLA
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