Analysis

7.1 Simple Functions And Measurable Functions

Recall that a σ algebra is a collection of subsets of a set Ω which includes ∅,Ω, and is closed with respect to countable unions and complements.

It is important to have a theorem about pointwise limits of measurable functions, those with the property that inverse images of open sets are measurable. The following is a fairly general such theorem which holds in the situations to be considered in these notes.

Proof: It is required to show f⁻¹

is measurable for all U open. Let

{ } ( C) 1- Vm ≡ x ∈ U : dist x,U > m .

Thus, since dist is continuous,

{ } V ⊆ x ∈ U : dist(x,UC) ≥ 1- m m

and V _m ⊆V _m ⊆ V _m+1 and ∪_mV _m = U. Then since V _m is open,

f− 1(Vm ) = ∪ ∞n=1 ∩∞k=n f−k1(Vm )

and so

which shows f⁻¹ is measurable. ■

An important example of a metric space is of course ℝ, ℂ, Fⁿ, where F is either ℝ or ℂ and so forth. However, it is also very convenient to consider the metric space (−∞,∞], the real line with ∞ tacked on at the end. This can be considered as a metric space in a very simple way.

ρ(x,y) = |arctan(x)− arctan (y)|

with the understanding that arctan

≡ π∕2. It is easy to show that this metric restricted to ℝ gives the same open sets on ℝ as the usual metric given by d = but in addition, allows the inclusion of that ideal point out at the end of the real line denoted as ∞. This is considered mainly because it makes the development of the theory easier. The open sets in (−∞,∞] are described in the following lemma.

Proof: If the center of the ball is a real number, then the ball will result in an interval

where a,b are real numbers. If the center of the ball is ∞, then the ball results in something of the form (a,∞]. It is obvious that this is a separable metric space with the countable dense set being ℚ since every ball contains a rational number. ■

If you kept both −∞ and ∞ with the obvious generalization that arctan

≡−, then the resulting metric space would be a complete separable metric space. However, it is not convenient to include −∞, so this won’t be done. The reason is that it will be desired to make sense of things like f + g.

Then for functions which have values in (−∞,∞] we have the following extremely useful description of what it means for a function to be measurable.

Proof: First note that the first and the third are equivalent. To see this, observe

−1 ∞ −1 f ([d,∞ ]) = ∩n=1f ((d− 1∕n,∞ ]),

and so if the first condition holds, then so does the third.

−1 ∞ − 1 f ((d,∞ ]) = ∪n=1f ([d+ 1∕n,∞ ]),

and so if the third condition holds, so does the first.

Similarly, the second and fourth conditions are equivalent. Now

−1 − 1 C f ((− ∞, d]) = (f ((d,∞ ]))

so the first and fourth conditions are equivalent. Thus the first four conditions are equivalent and if any of them hold, then for −∞ < a < b < ∞,

f−1((a,b)) = f−1((− ∞, b)) ∩f− 1((a,∞ ]) ∈ ℱ.

Finally, if the last condition holds,

f−1 ([d,∞ ]) = (∪ ∞k=1f −1((− k+ d,d)))C ∈ ℱ

and so the third condition holds. Therefore, all five conditions are equivalent.

Since (−∞,∞] is a separable metric space, it follows from Theorem 2.4.2 that every open set U is a countable union of open intervals U = ∪_kI_k where I_k is of the form

or (a,∞] and, as just shown if any of the equivalent conditions holds, then f⁻¹ = ∪_kf⁻¹ ∈ℱ. Conversely, if f⁻¹ ∈ℱ for any open set U ∈ (−∞,∞], then f⁻¹ ∈ℱ which is one of the equivalent conditions and so all the equivalent conditions hold.  ■

There is a fundamental theorem about the relationship of simple functions to measurable functions given in the next theorem.

Each simple function is measurable. This is easily seen as follows. First of all, you can assume the c_i are distinct because if not, you could just replace those E_i which correspond to a single value with their union. Then if you have any open interval

,

− 1 s ((a,b)) = ∪ {Ei : ci ∈ (a,b)}

and this is measurable because it is the finite union of measurable sets.

Proof: Letting I ≡

, define

∑2n k tn(ω) = nXf−1([kn,k+n1))(ω)+ 2nXI(ω). k=0

Then t_n(ω) ≤ f(ω) for all ω and lim_n→∞t_n(ω) = f(ω) for all ω. This is because t_n

= 2ⁿ for ω ∈ I and if f ∈ [0,), then

1 0 ≤ f (ω)− tn(ω) ≤-. (7.3) n

(7.3)

Thus whenever ω

I, the above inequality will hold for all n large enough. Let

s1 = t1,s2 = max (t1,t2),s3 = max (t1,t2,t3),⋅⋅⋅.

Then the sequence {s_n} satisfies 7.1-7.2. Also each s_n has finitely many values and is measurable. To see this, note that

s−1((a,∞]) = ∪n t−1((a,∞ ]) ∈ ℱ n k=1 k

To verify the last claim, note that in this case the term 2ⁿX_I(ω) is not present and for n large enough, 2ⁿ∕n is larger than all values of f. Therefore, for all n large enough, 7.3 holds for all ω. Thus the convergence is uniform.

The last claim follows right away from Theorem 7.1.2. ■

Although it is not needed here, there is a more general theorem which applies to measurable functions which have values in a separable metric space. In this context, a simple function is one which is of the form

∑m xkXEk (ω) k=1

where the E_k are disjoint measurable sets and the x_k are in X. I am abusing notation somewhat by using a sum. You can’t add in a general metric space. The symbol means the function has value x_k on the set E_k.

Proof: Let D =

_k=1^∞ be a countable dense subset of X. First suppose f is measurable. Then since in a metric space every open set is the countable intersection of closed sets, it follows f⁻¹ ∈ℱ. Now let D_n = _k=1ⁿ. Let

{ } A1 ≡ ω : d(x1,f (ω)) = min d(xk,f (ω)) k≤n

= ∩n {ω : d(x ,f (ω))− d(x ,f (ω)) ≥ 0} k=1 k 1

That is, A₁ is those ω such that f

is approximated best out of D_n by x₁. Why is this a measurable set? It is because ω → d − d is a real valued measurable function, being the composition of a continuous function, y → d − d and a measurable function, ω → f. Next let

{ } A ≡ ω∈∕A : d(x ,f (ω)) = min d(x ,f (ω)) 2 1 2 k≤n k

and continue in this manner obtaining disjoint measurable sets,

_k=1ⁿ such that for ω ∈ A_k the best approximation to f from D_n is x_k. Then

∑n fn(ω) ≡ xkXAk (ω). k=1

Note

min d (xk,f (ω)) ≤ mind (xk,f (ω )) k≤n+1 k≤n

and so this verifies 7.4. It remains to verify 7.5.

Let ε > 0 be given and pick ω ∈ Ω. Then there exists x_n ∈ D such that d

< ε. It follows from the construction that d ≤ d < ε. This proves the first half.

Now suppose the existence of the sequence of simple functions as described above. Each f_n is a measurable function because f_n⁻¹

= ∪. Therefore, the conclusion that f is measurable follows from Theorem 7.1.2 on Page 437. ■