304 CHAPTER 12. THE CONSTRUCTION OF MEASURES

Observe that from the definition,

lim infn→∞

1n

Sn f (ω) = lim infn→∞

1n

Sn f (T ω)

andlim sup

n→∞

1n

Sn f (ω) = lim supn→∞

1n

Sn f (T ω) .

Thus if ω ∈ Nab, it follows that T ω ∈ Nab and if T ω ∈ Nab, then so is ω . Thus Nab is aninvariant set. Also, if ω ∈ Nab, then

a− lim infn→∞

1n

Sn f (ω) = lim supn→∞

(a− 1

nSn f (ω)

)> 0

and

lim supn→∞

(1n

Sn f (ω)−b)> 0

It follows thatNab ⊆ [M∞ ( f −b)> 0]∩ [M∞ (a− f )> 0] .

Consequently, since Nab is invariant, argued above,

XNabM∞ ( f −b) = M∞

(XNab ( f −b)

)and so from Lemma 12.8.4∫

Nab

( f (ω)−b)dµ =∫[XNab M∞( f−b)>0]

XNab (ω)( f (ω)−b)dµ

=∫[M∞(XNab ( f−b))>0]

XNab (ω)( f (ω)−b)dµ ≥ 0 (12.8.27)

and ∫Nab

(a− f (ω))dµ =∫[XNab M∞(a− f )>0]

XNab (ω)(a− f (ω))dµ

=∫[M∞(XNab (a− f ))>0]

XNab (ω)(a− f (ω))dµ ≥ 0 (12.8.28)

It follows thataµ (Nab)≥

∫Nab

f dµ ≥ bµ (Nab) . (12.8.29)

Since a < b, it follows that µ (Nab) = 0.Now let

N ≡ ∪{Nab : a < b, a,b ∈Q} .

It follows that µ (N) = 0. Now T Na,b = Na,b and so

T (N) = ∪a,bT(Na,b

)= ∪a,bNa,b = N.