Question 1.2. Fix a measurable function \(f : \mathbb { R } ^ { 2 }\) R and, for every \(x , y \in \mathbb { R } ,\) , let
\[ f _ { x } : \mathbb { R } \to \mathbb { R } a n d f _ { y } : \mathbb { R } \to \mathbb { R } \]
be given by \(f _ { x } ( z ) = f ( x , z )\) and \(f _ { y } ( z ) = f ( z , y )\) . Show that there exists such an f so that \(f _ { x } \in L ^ { 1 } ( \mathbb { R } )\) for a.e. x and \(f _ { y } \in L ^ { 1 } ( \mathbb { R } )\) for a.e. y but
\[ \int _ { \mathbb { R } } \left( \int _ { \mathbb { R } } f _ { x } ( y ) d y \right) d x \neq \int _ { \mathbb { R } } \left( \int _ { \mathbb { R } } f _ { y } ( x ) d x \right) d y . \]
What does Fubini’s theorem imply about such \(f ?\) What about Tonelli’s theorem?
Solution
1. Define a measurable function with conditionally convergent iterated integrals.
Set \[ f(x,y)= \mathbf1_{(0,1)^2}(x,y) \frac{x^2-y^2}{(x^2+y^2)^2}, \] and define \(f=0\) on the coordinate axes. This is measurable on \(\mathbb R^2\).
For each fixed \(x\in(0,1)\), \[ \frac{x^2-y^2}{(x^2+y^2)^2} =\frac{d}{dy}\left(\frac{y}{x^2+y^2}\right), \] so \[ \int_{\mathbb R}f_x(y)\,dy =\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy =\frac1{1+x^2}. \] Thus \(f_x\in L^1(\mathbb R)\) for every \(x\ne0\).
For each fixed \(y\in(0,1)\), \[ \frac{x^2-y^2}{(x^2+y^2)^2} =-\frac{d}{dx}\left(\frac{x}{x^2+y^2}\right), \] hence \[ \int_{\mathbb R}f_y(x)\,dx =-\frac1{1+y^2}. \] Thus \(f_y\in L^1(\mathbb R)\) for every \(y\ne0\).
2. Compute the two iterated integrals.
Therefore \[ \int_{\mathbb R}\left(\int_{\mathbb R}f_x(y)\,dy\right)dx =\int_0^1\frac{dx}{1+x^2} =\frac\pi4, \] whereas \[ \int_{\mathbb R}\left(\int_{\mathbb R}f_y(x)\,dx\right)dy =-\int_0^1\frac{dy}{1+y^2} =-\frac\pi4. \] Hence the iterated integrals exist but are unequal.
3. Explain what Fubini and Tonelli imply.
Fubini’s theorem says that if \(f\in L^1(\mathbb R^2)\), then the two iterated integrals must agree. Since they do not, this example necessarily satisfies \[ f\notin L^1(\mathbb R^2). \]
This can also be seen directly from Tonelli’s theorem applied to \(|f|\). In polar coordinates in the first quadrant, \[ |f(r\cos\theta,r\sin\theta)| =\frac{|\cos(2\theta)|}{r^2}. \] On any angular subinterval where \(|\cos(2\theta)|\ge c>0\), the contribution near \(r=0\) is bounded below by a positive constant times \[ \int_0^\varepsilon\frac{dr}{r}=\infty. \] Hence \[ \iint_{\mathbb R^2}|f(x,y)|\,dx\,dy=\infty. \] So Tonelli confirms that absolute integrability fails, exactly as required for the two iterated integrals to differ.