Fubini–Tonelli Theorem

Inner integral first: for fixed $y$, $\int_0^1 xy\,dx = y\int_0^1 x\,dx = y \cdot \frac12$. The outer integral is $\int_0^1 \frac{y}{2}\,dy = \frac12 \cdot \frac12 = \frac14$.

Reversing the order: for fixed $x$, $\int_0^1 xy\,dy = x \cdot \frac12$, and $\int_0^1 \frac{x}{2}\,dx = \frac14$. Both orders give $\frac14$ — because the product splits: $\iint xy = (\int_0^1 x\,dx)(\int_0^1 y\,dy)$.

The function $f(x,y) = xy$ is continuous on $[0,1]^2$, hence measurable, and $f \ge 0$, so all hypotheses of Tonelli's theorem are satisfied and Tonelli's theorem guarantees the value is order-independent — the direct computation above verifies the theorem on this concrete function. Every step is complete.

Symmetry (the rigorous part): $f(y, x) = \dfrac{y^2-x^2}{(y^2+x^2)^2} = -\dfrac{x^2-y^2}{(x^2+y^2)^2} = -f(x, y)$.

Write the iterated integrals as $I_1 = \int_0^1\!\!\int_0^1 f(x,y)\,dy\,dx$ and $I_2 = \int_0^1\!\!\int_0^1 f(x,y)\,dx\,dy$. Renaming the dummy variables $x \leftrightarrow y$ in $I_2$ gives $I_2 = \int_0^1\!\!\int_0^1 f(y,x)\,dy\,dx = -\int_0^1\!\!\int_0^1 f(x,y)\,dy\,dx = -I_1$.

So the two iterated integrals are exact negatives of each other — proved with no computation.

An antiderivative in $y$ for $x > 0$: $\frac{\partial}{\partial y}\left[\frac{y}{x^2+y^2}\right] = \frac{(x^2+y^2) - y(2y)}{(x^2+y^2)^2} = \frac{x^2-y^2}{(x^2+y^2)^2} = f(x,y)$. Hence the inner integral $\int_0^1 f(x,y)\,dy = \left[\frac{y}{x^2+y^2}\right]_0^1 = \frac{1}{1+x^2} - 0 = \frac{1}{1+x^2}$.

At $x = 0$, $f(0,y) = \frac{-y^2}{y^4} = -\frac{1}{y^2}$, so the inner integral is improper: $\int_\varepsilon^1 -\frac{1}{y^2}\,dy = \left[\frac{1}{y}\right]_\varepsilon^1 = 1 - \frac{1}{\varepsilon} \to -\infty$ ($\varepsilon \to 0^+$). This happens only at the single point $x = 0$; the singleton $\{0\}$ has Lebesgue measure zero, so when the outer integral is read as a Lebesgue integral the inner integral only needs to exist for almost every $x$, and divergence at one point does not affect it: $\int_0^1 \frac{dx}{1+x^2} = [\arctan x]_0^1 = \frac{\pi}{4}$, hence $I_1 = \frac{\pi}{4}$.

Reversing the order, an antiderivative in $x$ is $\frac{\partial}{\partial x}\left[-\frac{x}{x^2+y^2}\right] = \frac{-(x^2+y^2) + x(2x)}{(x^2+y^2)^2} = \frac{x^2-y^2}{(x^2+y^2)^2} = f(x,y)$.

For $y > 0$, $\int_0^1 f(x,y)\,dx = \left[-\frac{x}{x^2+y^2}\right]_0^1 = -\frac{1}{1+y^2}$; at $y = 0$, $f(x,0) = \frac{1}{x^2}$, so $\int_\varepsilon^1 \frac{dx}{x^2} = \left[-\frac{1}{x}\right]_\varepsilon^1 = -1 + \frac{1}{\varepsilon} \to +\infty$ — again a one-point singularity.

The outer integral is $\int_0^1 -\frac{dy}{1+y^2} = -\frac{\pi}{4}$, so $I_2 = -\frac{\pi}{4}$.

Since $\frac{\pi}{4} \ne -\frac{\pi}{4}$, Fubini's conclusion fails. Before invoking the contrapositive, verify measurability and almost-everywhere inner integrability: $f$ is continuous on $(0,1]^2$, hence measurable on $[0,1]^2$ (define $f$ arbitrarily at the single point $(0,0)$; a single point and the coordinate segments $\{x=0\}$, $\{y=0\}$ have measure zero).

As shown above, the inner integrals exist for every $x > 0$ (resp. $y > 0$), i.e. for almost every $x$ (resp. $y$). Since $|f|$ is measurable and $\ge 0$, Tonelli applied to $|f|$ legitimately gives $\int |f| = \infty$.

Hence, by contrapositive, $f$ is not integrable over $(0,1)^2$ ($\int |f| = \infty$) — without the integrability hypothesis, signed mass cancels differently depending on the order.

The sign-flip symmetry was proved with no computation, and the $\pi/4$ evaluation is fully justified via antiderivatives and improper integrals. Every step is complete.

First, the indicator of a rectangle $R = A \times B$ (measurable sets, possibly of infinite measure): the inner integral $\int \mathbf{1}_R(x, y)\,dx = m(A)\,\mathbf{1}_B(y)$, and the outer gives $\int m(A)\,\mathbf{1}_B(y)\,dy = m(A)\,m(B)$.

Reversing the order, $\int \mathbf{1}_R(x, y)\,dy = m(B)\,\mathbf{1}_A(x)$ followed by $m(B)\,m(A)$ — the same.

Handle the degenerate case separately, with the measure-theoretic convention $0 \cdot \infty := 0$: if $m(B) = 0$ then $A \times B$ is null regardless of $A$, even if $m(A) = \infty$, so both iterated integrals are $0 = m(A)\,m(B)$; the case $m(A) = 0$ is symmetric.

For a simple function built from rectangles, $s = \sum_{p=1}^{P} c_p\,\mathbf{1}_{A_p \times B_p}$ ($c_p \ge 0$, disjoint rectangles), sections are explicit and no lemma is needed: for fixed $y$, $\int s(x,y)\,dx = \sum_p c_p\,m(A_p)\,\mathbf{1}_{B_p}(y)$ (previous step), a simple function of $y$, so the outer integral is $\sum_p c_p\,m(A_p)\,m(B_p)$ — the same with the order reversed.

Lemma (measurable sections): on a finite-measure product space (e.g. $[0,1]^2$), if $E$ is product-measurable then each section $E^y = \{x : (x,y) \in E\}$ is measurable, $y \mapsto m(E^y)$ is measurable, and $\int m(E^y)\,dy = m(E)$ (same for $x$-sections). Proof: first let $\mathcal{A}$ be the collection of finite disjoint unions of measurable rectangles; $\mathcal{A}$ is an algebra (closed under complements and finite unions) and generates the product $\sigma$-algebra, $\sigma(\mathcal{A})$ being the product $\sigma$-algebra.

The family $\mathcal{C}$ of sets with this property contains every rectangle — for $E = A \times B$, $E^y = A$ ($y \in B$) or $\varnothing$, and $m(E^y) = m(A)\,\mathbf{1}_B(y)$ — and by finite additivity the section-measure identity holds on all of $\mathcal{A}$, so $\mathcal{A} \subset \mathcal{C}$. Moreover $\mathcal{C}$ is a monotone class: if $E_n \uparrow E$ then $(E_n)^y \uparrow E^y$, so continuity of measure plus MCT gives $\int m((E_n)^y)\,dy = m(E_n) \uparrow m(E) = \int m(E^y)\,dy$ (decreasing sequences likewise, using finite measure).

By the monotone class theorem, $\mathcal{C}$ contains $\sigma(\mathcal{A})$, so $\mathcal{C}$ is the whole product $\sigma$-algebra.

For a general simple function $f = \sum_{k=1}^{m} a_k\,\mathbf{1}_{E_k}$ ($a_k \ge 0$, the $E_k$ disjoint product-measurable): the section $f^y(x) = f(x,y) = \sum_k a_k\,\mathbf{1}_{(E_k)^y}(x)$ is simple, so $\int f^y\,dx = \sum_k a_k\,m((E_k)^y)$ — and by the lemma, $y \mapsto \int f^y\,dx$ is a sum of measurable functions, hence measurable.

Integrating gives $\int\!\!\int f\,dx\,dy = \sum_k a_k \int m((E_k)^y)\,dy = \sum_{k=1}^{m} a_k\,m(E_k)$ — a single double sum, independent of order.

Every term is $\ge 0$, so nothing can cancel: no integrability hypothesis is needed.

For general $f \ge 0$, take the standard simple $s_j \uparrow f$ ($s_j = \sum_k a^{(j)}_k\,\mathbf{1}_{E^{(j)}_k}$).

For each fixed $y$, MCT in $x$ gives $\int s_j^y\,dx \uparrow \int f^y\,dx$ — and $\varphi_j(y) = \int s_j^y\,dx$ is an increasing sequence of measurable functions by the previous step, so one more MCT in $y$ gives $\int_y \varphi_j\,dy \uparrow \int_y(\int_x f\,dx)dy$.

But $\int_y \varphi_j\,dy = \sum_k a^{(j)}_k\,m(E^{(j)}_k) = \iint s_j \uparrow \iint f$ (MCT for the product measure), so both iterated integrals equal $\iint f$. This is why Tonelli's theorem needs no integrability hypothesis for $f \ge 0$ — every step is complete.