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Is there an injective and Riemann integrable map $f:\mathbb R^3\rightarrow\mathbb R$? (Of course such a map cannot be continuous.)

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Lebesgue integrable is not good enough? That would be easy. –  Goldstern Dec 30 '12 at 13:00
    
as long as it has a countable amount of discontinuities –  Ohad Asor Dec 30 '12 at 13:16
    
(which makes it Riemann) –  Ohad Asor Dec 30 '12 at 13:25
    
Riemann integrable makes sense where the domain is a bounded set. So if you really do want Riemann, perhaps you need to say what you mean. –  Gerald Edgar Dec 30 '12 at 13:32
    
ok. then Lebesgue with countable amount of discontinuities. thanks! –  Ohad Asor Dec 30 '12 at 13:39

2 Answers 2

up vote 6 down vote accepted

Consider for example $f \colon [0,1]^3 \to \mathbb{R}$ defined as $$f(x,y,z) = \sum_{i=0}^\infty 8^{-i}(x_i+2y_i+4z_i),$$ where $x_i = \left(\sum_{j=0}^{i-1}2^{i-j}x_j\right) - \lfloor x 2^i \rfloor$, and $y_i, z_i$ are defined the same way. In other words, $(x_i)_i$ is a specific dyadic representation of $x$.

Continuity: The mapping $f$ is continuous outside the union of the boundaries of dyadic cubes $$Q_{i,k_1,k_2,k_3} = 2^{-i}\left([k_1,k_1+1]\times [k_2,k_2+1] \times [k_3,k_3+1]\right),$$ for $i,k_1,k_2,k_3 \in \mathbb{Z}$. To see this, take $x \in [0,1]^3 \setminus \bigcup \partial Q_{i,k_1,k_2,k_3}$. Take $i \in \mathbb{N}$ and notice that for some $k_1,k_2,k_3 \in \mathbb{Z}$ (abbreviating $Q := Q_{i,k_1,k_2,k_3}$) we have $x \in \text{int}(Q)$. On the other hand $\text{osc}_f(Q) \le 8^{-i}$.

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This should work. But proof of continuity off a set of measure zero is needed. –  Gerald Edgar Dec 30 '12 at 20:40
    
Wonderful! How can you show that the integral over $\mathbb R^3$ is finite? In order to avoid infinities, can you calculate the integral of such a function but with domain $\mathbb S^n$ and range $[0,1]$? –  Ohad Asor Dec 31 '12 at 4:25
    
@Gerald: I added a few lines on the continuity outside the boundaries of unit cubes, who form a set of dimension 2. @Ohad: You can define $f$ on whole $\mathbb{R}^3$ by composing with translations and dilations like Gerald suggested. Since the original $f$ has bounded range, you can make the range to be inside your favorite bounded (non-empty and open) interval. Also, defining on $\mathbb{S}^n$ should be easy as well. –  Tapio Rajala Dec 31 '12 at 7:45
    
can you think of a way to evaluate this integral? –  Ohad Asor Dec 31 '12 at 12:00
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For the moment I can't think of any other way of estimating the integral except by direct calculation using the fact that you can write the integral explicitly as the sum of three geometric sums, giving $\int f = \frac{1}{2}(1+2+4)\sum_{i=1}^\infty 8^{-i} = \frac{1}{2}$. –  Tapio Rajala Dec 31 '12 at 14:40

beginning

Start with the unit cube $E$ in $\mathbb R^3$ and the unit interval $[0,1]$ in $\mathbb R$. Choose an injective map $\phi : E \to [0,1]$. This is the remaining question: do this so that $\phi$ is Riemann integrable. That is, the set of discontinuities has measure zero.

Then cover $\mathbb R^3$ by disjoint unit cubes, make them disjoint by including boundary points in only one of the possible cubes. Call these $F_n, n \in \mathbb N$. Choose disjoint intervals $I_n$ going to zero fast enough. Define $f$ using $\phi$ with translation and dilation to map $F_n$ injectively into $I_n$. If the original $\phi$ has set of discontinuities of measure zero, then this pieced-together function $f$ does too, since its set of discontinuities can be at most the discontinuities of a copies of $\phi$ in the interiors of the $F_n$, together with the boundary planes of the $F_n$. So $f$ is (improperly) Riemann integrable.

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