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
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 $x \in [0,1]$