21.8. THE RIESZ REPRESENTATION THEOREM 689

Then there is a sequence of simple functions {sn}which converge uniformly to f off a set ofmeasure zero, N, ∥sn∥L∞ ≤ ∥f∥L∞ . By regularity of the measure, there exists a continuousfunction with compact support hn such that sn = hn off a set of measure no more than δ/4n

and also ∥hn∥L∞ ≤ ∥f∥L∞ . Then off a set of measure no more than 13 δ , hn (r)→ f(r). Now

by Eggorov’s theorem and outer regularity, one can enlarge this exceptional set to obtainan open set S of measure no more than δ/2 such that the convergence is uniform off thisexceptional set. Thus f equals the uniform limit of continuous functions on SC. Define

h(r)≡

 limn→∞ hn (r) = f(r) on SC

0 on S\N0 on N

Then ∥h∥L∞ ≤ ∥f∥L∞ . Now consider

h̄∗ψm (r)

where ψr is approximate identity.

ψm (t) =12

mX[−1/m,1/m] (t) , h̄∗ψm (t) =12

m∫ 1/m

−1/mh̄(t− s)ds =

12

m∫ t+1/m

t−1/mh̄(s)ds

where we define h̄ to be the 0 extension of h̄ off [0,T ]. This is a continuous function of t.Also a.e.t is a Lebesgue point and so for a.e.t,∣∣∣∣12m

∫ t+1/m

t−1/mh̄(s)ds− h̄(t)

∣∣∣∣→ 0

∣∣h̄∗ψm (r)∣∣≡ ∣∣∣∣∫R h̄(r− s)ψm (s)ds

∣∣∣∣≤ ∥h∥L∞ ≤ ∥f∥L∞

Thus this continuous function is in L∞ (0,T,H). Letting z = zk ∈ L1 (0,T,H) be one ofthose defined above,∣∣∣∣∫ T

0

⟨h̄∗ψm (t)− f(t) ,z(t)

⟩dt∣∣∣∣≤ ∫ T

0

∣∣⟨h̄∗ψm (t)−h(t) ,z(t)⟩∣∣dt

+∫ T

0|⟨h(t)− f(t) ,z(t)⟩|dt (21.8.41)

for a.e. t, h̄∗ψm (t)− h(t)→ 0 and the integrand in the first integral is bounded by theexpression 2∥f∥L∞ |z(t)|H so by the dominated convergence theorem, as m→ ∞, the firstintegral converges to 0. As to the second, it is dominated by∫

S|⟨h(t)− f(t) ,z(t)⟩|dt ≤ 2∥f∥L∞

∫S|z(t)|dt <

2∥f∥L∞ ε

4(1+∥f∥L∞)≤ ε

2

Therefore, choosing m large enough so that the first integral on the right in 21.8.41 is lessthan ε

4 for each zk for k ≤M, then for each of these,

d(f, h̄∗ψm

)≤ ε

4+

M

∑k=1

2−k (ε/4)+(ε/2)1+((ε/4)+(ε/2))

=ε

4+

M

∑k=1

2−k 34

ε

34 ε +1

≤ ε

4+

3ε

4

M

∑k=1

2−k <ε

4+

3ε

4= ε