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In Fubini's theorem there's the function $f_y$, which is the "$y$-cut" of function $f$.

For $I=[a,b] \times [c,d]$ $$f_y:[a,b] \rightarrow \mathbb{R}, f_y(x)=f(x,y)$$

and then

$$F:[c,d] \rightarrow \mathbb{R}, F(y)=\int_a^b f_y(x)dx = \int_a^b f(x,y)dx$$

However, if $f_y(x)=f(x,y)$, then why is it a $y$-cut of $f$, rather than just function $f$?

Does $y$-cut mean that it's taken along the $x$-axis? Thus the $dx$.

mavavilj
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1 Answers1

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Because it's like a "section" or "cut" on the variable $y$. If you fix $y$, then $f_y(x)$ is a function that only depends on $x$. In the terms that you used, fixing an $y_0$ and moving the $x$ is moving over the line $y=y_0$ (parallel to the $X$-axis).

The same way you can define the $x$-cut: $f_x(y)=f(x,y)$.

A. Salguero-Alarcón
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