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

  • $\begingroup$ Set $x=y$ in the original equation? $\endgroup$ Nov 28, 2023 at 18:11
  • 1
    $\begingroup$ @Aruralreader The equation holds almost everywhere. $\endgroup$
    – Kroki
    Nov 28, 2023 at 18:12
  • $\begingroup$ 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. $\endgroup$
    – Yimin
    Nov 29, 2023 at 5:54

1 Answer 1


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 $x \in [0,1]$

  • $\begingroup$ I feel I might have been a little loose with claim (A) but my idea should be correct $\endgroup$
    – Snowflake
    Dec 1, 2023 at 6:43
  • $\begingroup$ @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 $\endgroup$
    – Snowflake
    Dec 3, 2023 at 14:15
  • $\begingroup$ @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 $\endgroup$
    – Snowflake
    Dec 3, 2023 at 14:16
  • $\begingroup$ 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. $\endgroup$ Dec 3, 2023 at 19:50
  • $\begingroup$ 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. $\endgroup$ Dec 6, 2023 at 3:27

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