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Let $g$ be a monotone function such that $$ \int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y)=0$$ where $f(x,y)= 1$ if $x-y \in \mathbb{Z}$ otherwise it is $0$.

The integration is w.r.t. Riemann-Stieltjes sense. How can we show that $g$ is continuous?

My try:

If $g$ is not continuous at $c$, $g$ can have only jump discontinuity at $c$ . I am not being able to use this fact. Any help or hint will be appreciated. Thanks in advance.

Update: If $f$ is not continuous at a point $c \in (0,1)$. Let us choose $\epsilon>0$ such that, $(c-\epsilon, c+ \epsilon) \subset (0,1)$. Then, $$ 0=\int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y) \geq \int_0^1 \int_0^1 f(x,x) \,dg(x) \,dg(x)= \int_0^1 \int_0^1 \,dg(x) \,dg(x) = [ \int_0^1 \,dg(x) ]^2 \geq [ \int_{c -\epsilon}^{c+ \epsilon} \,dg(x) ]^2 = [g(c+\epsilon)- g(c-\epsilon)]^2 >0$$, which is a contradiction. Thus $g$ is continuous at $c\in (0,1)$.

If $g$ is not continuous at $0$, then $$ 0=\int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y) \geq [g(\epsilon)- g(0)]^2 >0$$, a contradiction.

If $g$ is not continuous at $1$, then $$ 0=\int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y) \geq [ g(1)- g(1-\epsilon)]^2 >0$$, a contradiction.

Thus, $g$ is continuous on [0,1].

I want to justify the step $$ 0=\int_0^1 \int_0^1 f(x,y) \,dg(x) \,dg(y) \geq \int_0^1 \int_0^1 f(x,x) \,dg(x) \,dg(x)$$ with proof and hope the other parts are correct.

Any help or hint will be appreciated. Thanks in advance.

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  • $\begingroup$ Have you tried computing $\int_{0}^{1} f(x,y) dg(x)$? $\endgroup$ Dec 28 '20 at 1:28
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If $g$ is not continuous at some $0< c\le 1$ then, without loss of generality, $g(x)=I(x-c)$ so that $g$ has one and only one unit jump at $x=c.$

Now, $f=0$ except on the corners of the unit square and the diagonal, where $f=1.$ Then since $f(c,c)=1$, we see that $\int_0^1f(x,c)dI(x-c)$ does not exist, because $f$ has a discontinuity at the point where $g$ has a jump.

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    $\begingroup$ The continuity of $g$ does not imply $dg$ is absolutely continuous with respect to Lebesgue measure. The Cantor function is continuous, but $dg$ is singular. (Otherwise, I agree.) $\endgroup$ Dec 28 '20 at 18:32
  • $\begingroup$ Yes, indeed. Thank you. $\endgroup$ Dec 28 '20 at 20:17
  • $\begingroup$ Coukd you please elaborate little more? $\endgroup$ Dec 28 '20 at 21:42
  • $\begingroup$ On boundary $f=1?$ I think $f=1$ only on the diagonal and on $(0,1)$ and $(1,0)$ $\endgroup$ Dec 28 '20 at 21:45
  • $\begingroup$ How are you defining $g$ . $g$ is already given. Is not it? Kindly explain more. $\endgroup$ Dec 28 '20 at 21:46

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