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While waiting for responses to this question, I did some research and came across this function on $[0,1]^2$: $f(x,y) = 0$ if $x$ or $y$ is irrational and $f(x,y) = 1/q$ if $x$ and $y$ are rational and $x = p/q$ in lowest terms.

Its claimed that the double Riemann integral $\int_{[0,1]^2}f $ exists since $f$ is continuous almost everywhere, but if $x$ is rational then $\int_0^1 f(x,y)dy $ does not exist as a Riemann integral.

I understand the second part since $f(x,y)$ looks like the Dirichlet function (when $x =p/q$ fixed) alternating between $1/q$ and $0$ for rational and irrational $y$. Just because $f$ alternates between $0$ and a variable non-zero value off and on a rational grid does not make it completely obvious about the continuity.

So I would like to see how to prove the first part directly using Darboux sums.

RRL
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AlRacoon
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1 Answers1

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Clearly, for any partition $P$ of $[0,1]^2$, the lower sum satisfies $L(P,f) = 0$ since irrationals are dense.

Given $\epsilon >0$, choose a positive integer $N > 1/\epsilon$.

The set $A_N = \{x \in \mathbb{Q} \cap [0,1]: x = p/q, (p,q) = 1, q \leqslant N \}$ is finite. Here we have the rational numbers in $[0,1]$ where in lowest terms the denominator in $x = p/q$ is no bigger than $N$. Let $m = \#(A_N)$.

An upper sum can be split into a sum over subrectangles (1) including and (2) excluding points $(x,y)$ with $x \in A_N$:

$$U(P,f) = \sum_{(1)}M_j \, vol(R_j) + \sum_{(1)}M_j \, vol(R_j) $$

where $M_j$ is the supremum of $f$ over the subrectangle $R_j$.

For the first sum, there are at most $4m$ subrectangles $R_j$ including points where $f(x,y) = 1/q \geqslant 1/N$. Thus $1/N \leqslant M_j \leqslant 1$. Choosing a partition where the largest subrectangle has content less than $\epsilon/m$ we have

$$\sum_{(1)} \leqslant 4m \cdot 1 \cdot \sup_{R_j} vol(R_j) < 4\epsilon$$

For the second sum, the subrectangles contain no points with $x \in A_N$. Thus $M_j \leqslant 1/N < \epsilon$ and

$$\sum_{(2)} \leqslant \epsilon \sum_{(2)} vol(R_j) < \epsilon.$$

Therefore, there is a partition $P$ such that $U(P,f) < 5\epsilon$ proving that $f$ is Riemann integrable and

$$\int_{[0,1]^2} f = 0$$

RRL
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  • Dear RRL, could you please clarify why there are at most $4m$ subrectangles $R_j$ with the property you mentioned and how the inequalities in the next sentence follow? – user557 Feb 27 '18 at 00:13
  • @user531299: We have chosen the partition so that the content (area) of the largest subrectangle is less than $\epsilon/m$. The m points where $f(x,y) \geqslant 1/N$ are distributed among the subrectangles of the partition. To be safe, the number $4m$ is an overestimate of how many subrectangles contain the points. One point could belong to $0,1,2$ or $4$ rectangles -- where it is $4$ if it so happens that the point is a common vertex shared by $4$ rectangles. All we really need is that $O(m)$ rectangles contain the points. – RRL Feb 27 '18 at 04:44
  • The estimate of the sum in the next line is the maximum possible number of subrectangles of type (1) -- $4m$ as an overestimate-- times the largest value $f$ can take -- 1 since this is a global upper bound for the function -- times the largest possible area of a subrectangle $\epsilon/m$. This equals $4m \cdot 1\cdot \epsilon/m = 4\epsilon$. – RRL Feb 27 '18 at 04:48