 Solution

ID: past-exam-of-the-martingales-and-financial-mathematics-course-of-the-university-of-oxford/2007/a/1/b/iii/solution

The two child intervals partition their parent:

B_{n, k} = B_{n + 1, 2 k} \dot{\cup} B_{n + 1, 2 k + 1} .

Their preimages under

g

satisfy the same identity. Since all

A

sets lie inside

f (M)

and

f

is onto its image, equality of their preimages under

f

implies

A_{n, k} = A_{n + 1, 2 k} \dot{\cup} A_{n + 1, 2 k + 1} .

h_{n + 1} = 2 k 2^{- n - 1} = k 2^{- n} = h_{n}

; on the second it is

h_{n} + 2^{- n - 1}

. Outside the image both functions are zero. Thus

0 \leq h_{n + 1} (t) - h_{n} (t) \leq 2^{- n - 1} .

For

m > n

, summing these bounds gives

∣ h_{m} (t) - h_{n} (t) ∣ \leq \sum_{j = n}^{m - 1} 2^{- j - 1} \leq 2^{- n} .

This tends to zero uniformly in

m > n

, proving that each real sequence

h_{n} (t)

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