# Function whose graph is dense in the plane [duplicate]

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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• Maybe here you can found something interesting arxiv.org/abs/1404.5876 – Tito Eliatron Jan 15 at 21:49
• – copper.hat Jan 15 at 21:50
• "Covers the plane" is not the right title. You want a function whose GRAPH is dense in the plane. I'll edit for you. – zhw. Jan 15 at 22:27
• @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. – BigbearZzz Jan 16 at 10:54

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