Kenneth Kuttler

10.1. LINEAR TRANSFORMATIONS 241

since this transformation just magnifies one side, multiplying it by c.The other linear transformation which represents a sheer is a little harder. However,

mp (E (s→ s+ r)(R)) =∫

E(s→s+r)(R)dmp

=∫R· · ·∫R

∫R

XE(s→s+r)(R)dxsdxrdxp1 · · ·dxpp−2

Now recall Theorem 7.8.5 which says you can integrate using the usual Riemann inte-gral when the function involved is Borel. Thus the above becomes∫ bpp−2

app−2

· · ·∫ bp1

ap1

∫ br

∫ bs+xr

as+xr

dxsdxrdxp1 · · ·dxpp−2

= mp (R) = |det(E (s→ s+ r))|mp (R)

Recall that when a row (column) is added to another row (column), the determinant of theresulting matrix is unchanged.

Lemma 10.1.3 Let L be any of the above elementary linear transformations. Then

mp (L(F)) = |det(L)|mp (F)

for any Borel set F. Also L(F) is Lebesgue measurable if F is Lebesgue measurable. If Fis Borel, then so is L(F).

Proof: Let Rk = ∏pi=1 (−k,k) . Let G be those Borel sets F such that L(F) is Borel and

mp (L(F ∩Rk)) = |det(L)|mp (F ∩Rk) (10.1)

Letting K be the open rectangles, it follows from the above discussion that the pi systemK is in G . It is also obvious that if Fi ∈ G the Fi being disjoint, then since L is one to one,

mp (L(∪∞i=1Fi∩Rk)) =

∞

∑i=1

mp (L(Fi∩Rk)) = |det(L)|∞

∑i=1

mp (Fi∩Rk)

= |det(L)|mp (∪∞i=1Fi∩Rk)

Thus G is closed with respect to countable disjoint unions. If F ∈ G then

mp(L(FC ∩Rk

))+mp (L(F ∩Rk)) = mp (L(Rk))

mp(L(FC ∩Rk

))+ |det(L)|mp (F ∩Rk) = |det(L)|mp (Rk)

mp(L(FC ∩Rk

))= |det(L)|mp (Rk)−|det(L)|mp (F ∩Rk)

= |det(L)|mp(FC ∩Rk

)It follows that G is closed with respect to complements also. Therefore, G = σ (K ) =B (Rp). Now let k→ ∞ in 10.1 to obtain the desired conclusion. ■

10.1. LINEAR TRANSFORMATIONS 241since this transformation just magnifies one side, multiplying it by c.The other linear transformation which represents a sheer is a little harder. However,my (E(s—s+r)(R)) = I comm)(R)~ Le LAL BE (ssstr)(R)AXsdXrdXp, *--AXpy_5Now recall Theorem 7.8.5 which says you can integrate using the usual Riemann inte-gral when the function involved is Borel. Thus the above becomesbpy_o bp, por phstxrtee dxsdX,dXp, “+AdXp, 55 ap y_2 Jap, Jar SastXr= mp(R) = |det(E(s > s+r))|mp(R)Recall that when a row (column) is added to another row (column), the determinant of theresulting matrix is unchanged.Lemma 10.1.3 Let L be any of the above elementary linear transformations. ThenMp (L(F)) = |det (L)| mp (F)for any Borel set F. Also L(F) is Lebesgue measurable if F is Lebesgue measurable. If Fis Borel, then so is L(F).Proof: Let Ry = []}_, (—k,k) . Let Y be those Borel sets F such that L(F) is Borel andmp (L(F A Rx)) = |det (L)| mp (F A Rx) (10.1)Letting .% be the open rectangles, it follows from the above discussion that the pi systemH isin Y. It is also obvious that if F; € Y the F; being disjoint, then since L is one to one,mp (L(URFR:)) = Limp (L(RORk)) = [det (L)| mp (ORs)i=l i=l= |det (L)| mp (Uj=1 FiO Re)Thus Y is closed with respect to countable disjoint unions. If F € Y thenmy (L(F©ORx)) + mp (L(F Rg) = mp (L(Re))my (L (FO ARx)) + [det (L)| mp (F ARx) = |det (L)| mp (Rx)my (L(F©ORx)) |det (L)| mp (Rx) — |det (L)| mp (FO Rx)|det (L)| mp (F° NR)It follows that Y is closed with respect to complements also. Therefore, Y = 0 (.%) =&(R?’). Now let k — in 10.1 to obtain the desired conclusion. Hi