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This question already has an answer here:

Is there a function $f:\mathbb R\to\mathbb R$ such that for every disc in $\mathbb R^2$ the graph of that function has at least one point that lies inside that disc? I searched for something similar but don't know which phrases to use. The general/another problem is that: does there exist a function $f:\mathbb R\to\mathbb R$ such that its graph in some sense has a positive area or even an infinite area?

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marked as duplicate by copper.hat real-analysis Jan 15 at 23:44

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Maybe here you can found something interesting arxiv.org/abs/1404.5876 $\endgroup$ – Tito Eliatron Jan 15 at 21:49
  • $\begingroup$ Go linear math.stackexchange.com/q/227427/27978 $\endgroup$ – copper.hat Jan 15 at 21:50
  • $\begingroup$ "Covers the plane" is not the right title. You want a function whose GRAPH is dense in the plane. I'll edit for you. $\endgroup$ – zhw. Jan 15 at 22:27
  • $\begingroup$ @copper.hat I don't think this question is an exact duplicate of the one you linked to because this one also ask whether such graph has positive area. $\endgroup$ – BigbearZzz Jan 16 at 10:54
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As mentioned earlier in the comment, the graph of a discontinuous linear map is dense so this answer your first question.

However, a graph of a function $f:\Bbb R\to\Bbb R$ cannot have positive area. This can be deduced from Fubini's theorem, for example, if we interpret the word "area" to mean the (two-dimensional) Lebesgue measure of $\text{Gr}(f)=\{(x,f(x)) : x\in\Bbb R\}$. Indeed, let $E=\text{Gr}(f)$ then $$\begin{align} \mu(E) &= \int_{\Bbb R^2} \chi_{E} d\mu \\ &= \int_{\Bbb R}\int_{\Bbb R} \chi_E(x,y) \,dy\,dx \end{align}$$ but for each $x\in\Bbb R$ we know that $$ \int_{\Bbb R} \chi_E(x,y) \,dy = 0 $$ because $\chi_E(x,y)=0$ almost everywhere except the point $y=f(x)$, which doesn't effect the value of the integral. Hence we conclude that $\mu(E)=0 $ so $\text{Gr}(f)$ has zero area.


Remark: The situation is different though if you consider from the start a function $f:\Bbb R\to \Bbb R^2$. The meaning of "graph of $f$" would also be different here: it would merely be the image of $\Bbb R$ under $f$, i.e. $$ G=\{p\in\Bbb R^2: p=f(x) \text{ for some }x\in \Bbb R \}. $$ In this case there are many known space-filling curves that have positive area in $\Bbb R^2$. The most well-known is probably Peano's curve.

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I do not know if $g$ as you wish exists, it is very odd if there exists, but there is a curve called "Piano's curve" described as such: $\gamma :[0,1]\to [0,1]^2$ s.t $\forall p\in [0,1]^2\, \,\exists t\in[0,1] : \gamma (t)=p$, i.e, $\gamma$ is space filling curve. you can obviously extend it to any closed square by shifting and rescaling. I think it answers your last question, this curve has area of 1.

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  • $\begingroup$ I wondered about a function with the codomain $\mathbb R$ however. $\endgroup$ – M. Świderski Jan 16 at 17:27

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