= Solution
A <lambda-system> contains the whole space and is closed under complements and countable disjoint unions. Let $\mathcal D$ be the smallest lambda-system containing a <pi-system> $\mathcal P$. For $A\in\mathcal P$, the family of $B\in\mathcal D$ with $A\cap B\in\mathcal D$ is a lambda-system containing $\mathcal P$, hence is all of $\mathcal D$. Fixing $B\in\mathcal D$ and repeating the argument shows $\mathcal D$ is closed under intersections. A lambda-system closed under intersections is a <sigma-algebra>, so $\mathcal D=\sigma(\mathcal P)$. This proves the <pi-lambda theorem>. If two <probability measures> agree on $\mathcal P$, the sets on which they agree form a lambda-system; thus they agree on $\sigma(\mathcal P)$.
The nonnegative form of <Fubini's theorem>, also called <Tonelli theorem>, states that for a nonnegative product-measurable function on two sigma-finite measure spaces, the product integral equals either iterated integral, allowing the value infinity.
For <Lebesgue measure> on the plane, a bounded interval rectangle $I\times J$ has measure $|I||J|$. Its preimage under $f$ is $(\lambda I)\times(\lambda^{-1}J)$, which has the same measure. Under $g$, its vertical slice at $x\in I$ is $J-sx$, whose one-dimensional Lebesgue measure is $|J|$ by translation invariance. Tonelli therefore gives $\mu(g^{-1}(I\times J))=|I||J|$. Horizontal slices likewise prove the equality for $h$. The interval rectangles form a generating pi-system. Sigma-finite measure uniqueness, obtained by the finite-measure pi-lambda argument on an increasing rectangular exhaustion, extends these equalities to all Borel sets. Completion extends them to all Lebesgue-measurable sets, since Borel null sets and their subsets remain null.
Writing $\lambda^2=c$, matrix multiplication gives
$$
f h g f=\begin{pmatrix}c&-s\\s&c\end{pmatrix}.
$$
Thus rotations through angles in $[0,\pi/2)$ preserve measure. Their inverses and compositions do too; every plane rotation is a finite composition of these, including a right angle as two rotations through $\pi/4$. \b[Lebesgue measure is invariant under every rotation.]
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