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As far as I know, the essential supremum of a measurable function $f:[0,1]\to\mathbb R$ is defined as the infimum of all essential upper bounds. Equivalently, \begin{align*} \text{ess}\sup f=\inf\{\sup f([0,1]\backslash N)\ |\ N\subset[0,1],|N|=0\}, \end{align*} where $|N|$ denotes the Lebesgue-measure. Similar for $\text{ess}\inf$.

From this it follows that if $f:[0,1]\to\mathbb R$ is a measurable Darboux-function (Darboux-function means, that for each $a,b\in[0,1]$ and each $y$ between $f(a)$ and $f(b)$ there is some $x$ between $a$ and $b$ with $f(x)=y$), then \begin{align*} \text{ess}\inf f=\inf f\qquad \text{and}\qquad \text{ess}\sup f=\sup f. \end{align*} To see this, define $c:=\inf f$ and $d:=\sup f$. Then $f([0,1])\subseteq[c,d]$. Fix $y\in(c,d)$. Then one can find $a,b\in[0,1]$ such that \begin{align*} c\leq f(a)\leq y\leq f(b)\leq d. \end{align*} Since $f$ is Darboux, there is some $x\in[a,b]$ (or $x\in[b,a]$) such that $f(x)=y$. This shows $(c,d)\subseteq f([0,1])$, and in total, \begin{align*} (c,d)\subseteq f([0,1])\subseteq[c,d]. \end{align*} But then indeed $\text{ess}\inf f=\inf f$ and $\text{ess}\sup f=\sup f$.

From this it follows that if $f:[0,1]\to\mathbb R$ is a measurable Darboux-function which is zero almost everywhere, then $f$ is identically zero. But this contradicts this article.

Where is my mistake?

Thank you in advance! I must be overlooking something trivial...

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  • $\begingroup$ Well, if $c=\inf f$ and $d=\sup f$, then $c\leq f(x)\leq d$ for each $x\in[0,1]$, which in short is nothing but $f([0,1])\subseteq[c,d]$... $\endgroup$
    – sranthrop
    Commented Aug 23, 2017 at 0:55
  • $\begingroup$ I don't know that and it doesn't matter. Even if $c=-\infty$ and $d=\infty$, then $f([0,1])=\mathbb R$ and $[c,d]=\mathbb R\cup\{-\infty,\infty\}$, and the inequality is still true. Am I overlooking something? $\endgroup$
    – sranthrop
    Commented Aug 23, 2017 at 1:08
  • $\begingroup$ Btw: As far as I can tell, the function constructed in this article is non-negative and bounded from above (since it is upper-semicontinuous). $\endgroup$
    – sranthrop
    Commented Aug 23, 2017 at 1:14

2 Answers 2

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If you're interested, here is a way to see there are Darboux functions equal to $0$ a.e. without being identically $0:$ Let $I_n, n = 1,2,\dots$ be the open intervals with rational end points. We can inductively choose pairwise disjoint Cantor sets $K_n\subset I_n$ with $m(K_n)=0$ for all $n.$

Now each $K_n$ has the cardinality of $\mathbb R.$ Hence there are bijections $f_n: K_n \to \mathbb R, n=1,2,\dots$ Define $f:\mathbb R\to \mathbb R$ by setting $f=f_n$ on each $K_n,$ $f=0$ everywhere else. Then $f=0$ on $\mathbb R\setminus (\cup K_n).$ Since $K_n$ has measure $0,$ so does $\cup K_n.$ Hence $f=0$ a.e.

Let $a<b.$ Then there exist some $I_n \subset (a,b),$ which implies $K_n \subset (a,b).$ Since $f$ maps $K_n$ onto $\mathbb R,$ we have $f((a,b))= \mathbb R,$ and the Darboux property of $f$ follows.

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  • $\begingroup$ Oh that's really beatiful. Thank you very much! $\endgroup$
    – sranthrop
    Commented Aug 25, 2017 at 9:41
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The conclusion that $(c,d)\subset f([0,1])$ implies $d=\text{ess}\sup (c,d)\leq\text{ess}\sup f$ is wrong. The reason for this is that $f$ can attain its image set on a nullset and be something completely different on all other points. If one removes this nullset to calculate the essential supremum, $f$ can have values less than and bounded away from $d$, leading to an essential supremum that is strictly less than $d$. The same is true for the infimum.

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