# Integration is zero implies $g$ is continuous

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.

• Have you tried computing $\int_{0}^{1} f(x,y) dg(x)$? Dec 28 '20 at 1:28

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.
• 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.) Dec 28 '20 at 18:32
• On boundary $f=1?$ I think $f=1$ only on the diagonal and on $(0,1)$ and $(1,0)$ Dec 28 '20 at 21:45
• How are you defining $g$ . $g$ is already given. Is not it? Kindly explain more. Dec 28 '20 at 21:46