# integral evaluating to a quantity 'almost everywhere' in $\mathbb{R}^2$

Let $$g:[0,1]^2 \to [0,1]$$ be a measurable function. Suppose $$\int\limits_0^1 g(x,t)g(y,t)dt = A$$ holds for almost every $$(x,y)\in [0,1]^2$$ then prove that $$\int\limits_0^1 g^2(x,t)dt = A$$ for almost every $$x \in [0,1]$$

I found the question in an assignment so I am not sure if the question has a typo or not but I am led to believe so. Suppose we have that the condition in the hypothesis does not hold over the line segment $$x=y$$ lying inside the $$2$$ dimensional unit square (which of course has measure zero in $$[0,1]^2$$). Then clearly the conclusion will not follow, right?

I need help finding a counter example OR if my intuition is wrong, I would like to see some ideas towards a proof

• Set $x=y$ in the original equation? Nov 28, 2023 at 18:11
• @Aruralreader The equation holds almost everywhere. Nov 28, 2023 at 18:12
• if there is a positive measure set $S$ that $\int g^2(x,t) dt < A - \epsilon$, then by AM-GM, on $S\times S$, there is a contradiction. It seems there is no control on the other side. Nov 29, 2023 at 5:54

## 1 Answer

We prove two facts after giving some definitions.

Let $$g_x$$ denote the function $$g_x(y)=g(x,y)$$ and let $$f \in L^2([0,1])$$ be another function that is in the $$L^2$$ space over $$[0,1]$$

We define $$N_{\varepsilon}(f)= \{x\in[0,1] : \|{f-g_x}\|_2<\varepsilon\}$$

We also define $$\mathcal{T}=\{f \in L^2([0,1]) : m\left( N_{\varepsilon}(f)\right) > 0 \hspace{0.2cm}\forall \hspace{0.15cm} \varepsilon >0 \}$$ where $$m$$ denotes the lesbegue measure

Let $$S\subseteq L^2([0,1])$$ be a countable dense subset and let $$B=\bigcup\limits_{\substack{f\in S \\ \varepsilon \in \mathbb{Q}_{+} \\ m\left( N_{\varepsilon}(f)\right) = 0}} N_{\varepsilon}(f)$$

Claim A: For all $$f,g\in\mathcal{T}$$ we have that $$\int_{[0,1]} fg=A$$

Claim B: If $$g_x\notin \mathcal{T}$$, then $$x\in B$$

Proof of claims -

For claim (A):

We know that $$f,g\in \mathcal{T}$$ and hence, we have that, for any given $$\varepsilon >0$$,

$$\int_{[0,1]} |f(z)-g(x,z)|dz < \varepsilon$$ and $$\int_{[0,1]} |g(z)-g(x,z)|dz < \varepsilon$$ hold almost everywhere

Consider $$\int_{[0,1]} |f(z)g(z)-g(x,z)g(x,z)|dz \leq \int_{[0,1]} |f(z)||g(z)-g(x,z)|dz + \int_{[0,1]} |g(x,z)||f(z)-g(x,z)|dz$$

Now, $$\int_{[0,1]} |f(z)||g(z)-g(x,z)|dz \leq M\varepsilon$$ almost everywhere and $$\int_{[0,1]} |g(x,z)||f(z)-g(x,z)|dz \leq \varepsilon$$ again, almost everywhere

where we used the fact that $$M = \int_{[0,1]} f(z)dz$$ (exists since $$f\in L^2([0,1])$$) and $$|g(x,z)|\leq 1$$ everywhere by definition

Since $$\varepsilon$$ is arbitrary and $$\varepsilon > 0$$, we have that our integral we considered is actually zero and hence $$\int_{[0,1]} f(z)g(z)dz = \int_{[0,1]} g(x,z)g(x,z)dz = A$$

For claim B:

Let for some $$x_0 \in [0,1], g_{x_0} \notin \mathcal{T}$$.

Thus there exists some $$\varepsilon_0$$ so that $$N_{\varepsilon_0}(g_{x_0})$$ has measure $$0$$.

Fix $$\delta=\min\limits_{x\in[0,1]\backslash {x_0}} \{ \| g_{x_0}-g_x\|_2\}$$.

Clearly $$\delta >0$$

Choose $$h$$ from $$S$$ such that $$g_{x_0}$$ and $$h$$ are less than $$\delta$$ close to each other.

Such an $$h$$ exists due to the density of $$S$$

We claim that $$N_{\delta}(h)$$ has measure zero.

Supposing our claim is true, we see that $$x_0 \in N_{\delta}(h)$$ by choice(definition) of $$h$$

Now the claim is true. This is seen in the following idea.

Consider the measure $$m\left( \{ x \in [0,1] : \| h- g_x \| < \delta \}\right)$$

This will have to be zero since, if there is any $$x$$ other than $$x_0$$ present in the given set, it will lead to a contradiction by definition of $$\delta$$

Thus, we are done since we have found an $$f$$ and $$\delta$$ such that $$m(N_{\delta}(f))=0$$ and $$x_0 \in N_{\delta}(f)$$

Note that $$\delta$$ can be tuned accordingly to become a rational number since the set of rationals is dense in $$[0,1]$$

Proof of actual question I had asked using our claims:

By claim (A), if $$g_x \in \mathcal{T}$$ then we have that $$\int\limits_0^1 (g_x(z))^2 dz$$

However if $$g_x \notin \mathcal{T}$$ then we have that $$x$$ belongs to a countable union of sets that have measure zero by claim (B) and hence $$x$$ is in a set of measure zero (countable union of sets of measure zero has measure zero)

Thus, we have that the conclusion holds everywhere except at points which form together a set of measure zero and hence we can say that the conclusion holds for almost all $$x \in [0,1]$$

• I feel I might have been a little loose with claim (A) but my idea should be correct Dec 1, 2023 at 6:43
• @EliJohnson Nope ! For the $g$ you provided, it does not even satisfy the initial hypothesis of integral of $g_x \cdot g_y$ being equal to $A$ almost everywhere Dec 3, 2023 at 14:15
• @EliJohnson Nope ! For the $g$ you provided, it does not even satisfy the initial hypothesis of integral of $g_x \cdot g_y$ being equal to $A$ almost everywhere Dec 3, 2023 at 14:16
• Fair point, I'll delete! But I'm not able to justify your statement that $f\in\mathcal{T}$ implies $\|f-g_x\|_{L^1([0,1])}<\varepsilon$ almost everywhere. You've only shown this is guaranteed true on $N_\varepsilon(f)$, i.e. a set of nonzero measure. Also, in your proof of claim B, I don't agree you've shown $\delta>0$. Take e.g. $g(x,t) = 1$ when $x\in \mathbb{Q}$ and $=0$ otherwise. In that case $\delta = 0$. A minor issue is that your reuse of $g$ as an arbitrary function in $\mathcal{T}$ is a bit confusing as it relates to the function of the hypothesis. Dec 3, 2023 at 19:50
• I am seeing another issue. You haven't shown $\mathcal{T}$ is nonempty. It is not true in general that $\{x:g_x\in\mathcal{T}\} \cup B = [1,0]$. Take $g(x,t)$ to be any function that is never essentially constant in $x$. More precisely, take $g$ so that there is no positive measure set $E\subseteq [0,1]$ such that $x,y\in E \implies g_x = g_y$ a.e. in $t$. Then $\mathcal{T}=\varnothing$, even while your $B$ remains a null set. Intuitively, it seems far-fetched that such a function $g$ could satisfy the hypothesis, but you haven't ruled it out. Dec 6, 2023 at 3:27