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I want an example of a measurable function $f:[0,1]\rightarrow [0,1]$ such that

$(1)$ $f$ is strictly increasing almost every where,

$(2)$ Upper Riemann integral of $f$is $1$,

$(3)$ Lower Riemann integral of $f$ is $0$

$(4)$ $0<\int_{[0,1]}fdm <1 $ where $m$ is Lebesgue measure. Can any one give Hints?

Math
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    What do you mean by "$f$ is strictly increasing almost everywhere"? Is that to say that there is a set $A\subset [0,1]$ of measure $1$ such that the restriction of $f$ to $A$ is an increasing function? – Ben Grossmann Mar 25 '14 at 11:54
  • Omnomnomnom : "f" is strictly increasing almost every where means that the set of points where f is not strictly increasing has measure zero. – Math Mar 25 '14 at 11:57
  • "$f$ is increasing at a point" is not a meaningful statement. Or, if you mean something by it, you'll have to clarify what it is that you mean. – Ben Grossmann Mar 25 '14 at 11:58
  • @Omnomnomnom You could interpret "f is strictly increasing at $x$" as "There is an $\epsilon$ such that $\forall y \in (x-\epsilon,x) ,:, f(y) < f(x)$ and $\forall y \in (x,x+\epsilon) ,:, f(x) < f(y)$". Not sure how usefull that definition is, though. – fgp Mar 25 '14 at 12:08
  • @fgp for all I know, that could be what is really meant. Importantly, that definition would preclude the existence of a counterexample. – Ben Grossmann Mar 25 '14 at 12:11
  • @Math is there a context for this problem? Is this how the problem is stated? Again, your definition of "strictly increasing almost everywhere" matters, and the definition that you've given is meaningless. – Ben Grossmann Mar 25 '14 at 12:14
  • Omnomnomnom:As to the concept of almost every where()a.e f" is strictly increasing almost every where means that the set of points where f is not increasing has measure zero. – Math Mar 25 '14 at 12:22
  • @Math what do you mean by $f$ is increasing at a point??? – Ben Grossmann Mar 25 '14 at 12:28
  • @Math: A function $f$ is increasing on a set $A \subset \mathbb{R}$ if for any $x,y \in A$, we have $f(x) > f(y)$ whenever $x>y$. We say that $f$ is increasing at a point $x$ if... ???? – Ben Grossmann Mar 25 '14 at 13:27
  • Note that if this is how the question was originally phrased, we're probably using the wrong definition. – Ben Grossmann Mar 25 '14 at 17:40

2 Answers2

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Let $E \subset Q^c$ be a countable dense subset in $[0,1]$. Define

$f(x) : = \begin{cases} 0 \mbox{ if } x \in \mathbb{Q} \cap [0,1]\\ 1 \mbox{ if } x \in E\\ x \mbox{ otherwise } \end{cases}$

DiffeoR
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An example of what I think you mean: define $A_1 = \mathbb{Q}\cap [0,1]$, $A_2 = \{\sqrt{2}q:q \in \mathbb{Q} \cap (0,1]\}$, and $A_3 = [0,1]\setminus (A_1 \cup A_2)$. Define $$ f(x) = \begin{cases} 0 & x \in A_1\\ 1 & x \in A_2\\ x & x \in A_3 \end{cases} $$

Ben Grossmann
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