Does the Lebesgue measure on the segment $y=x$ represent this distribution?

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Set $\Omega=(-1,1)^2$. Consider the following measure on $\Omega$:

$\mu(A)=m(A \cap L)$, where $L=\{ (x,x) \, | \, -1 < x < 1\}$ (the segment of the line $y=x$ in $\Omega$) , and $m$ is the Lebesgue measure on $L$ (or the corresponding Hausdorff measure, I guess).

Let $\phi$ be a smooth compactly supported real-valued function on $\Omega$.

Is it true that $\int_{\Omega} \phi d\mu=\int_{-1}^1 \phi(x,x)dx$? If not, then what (signed) measure does represent the distribution $\phi \to \int_{-1}^1 \phi(x,x)dx$?

(I think there might be some subtle issue with the orientation, as I don't see how the measure $\mu$ "knows" that the right direction is $-1 \to 1$ and not $1 \to -1$. That is, it does not seem that it encodes this information in any way).

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It seems like $\mu$ is the push forward of Lebesgue measure on (-1,1) by the map $x \rightarrow (x, x) $, maybe up to a $\sqrt{2}$ factor. Then the equation you want is exactly the definition of the push-forward measure. Am I missing something or is this correct?