tions
The Monotone Covergence theorem is one of a number of key theorems alllowing one to ex- change limits and [Lebesgue] integrals (or derivatives and integrals, as derivatives are also a sort of limit). Fatou’s lemma and the dominated convergence theorem are other theorems in this vein, where monotonicity is not required but something else is needed in its place. In Fatou’s lemma we get only an inequality for lim inf’s and non-negative integrands, while in the dominated con- vergence theorem we can manage integrands that change sign but we need a ‘dominating’ inte- grable function as well as existence of pointwise limits of the sequence of inetgrands.
Contents
4.1 Fatou’s Lemma . . . 1
4.2 Almost everywhere . . . 2
4.3 Dominated convergence theorem . . . 4
4.4 Applications of the dominated convergence theorem . . . 5
4.5 What’s missing? . . . 8
4.1 Fatou’s Lemma
This deals with non-negative functions only but we get away from monotone sequences.
Theorem 4.1.1 (Fatou’s Lemma). Let fn: R → [0, ∞] be (nonnegative) Lebesgue measurable functions. Then
lim inf
n→∞
Z
R
fndµ ≥ Z
R
lim inf
n→∞ fndµ
Proof. Let gn(x) = infk≥nfk(x) so that what we mean by lim infn→∞fn is the function with value at x ∈ R given by
lim inf
n→∞ fn
(x) = lim inf
n→∞ fn(x) = lim
n→∞
k≥ninf fk(x)
= lim
n→∞gn(x)
Notice that gn(x) = infk≥nfk(x) ≤ infk≥n+1fk(x) = gn+1(x) so that the sequence (gn(x))∞n=1 is monotone increasing for each x and so the Monotone convergence theorem says that
n→∞lim Z
R
gndx = Z
R
n→∞lim gndµ = Z
R
lim inf
n→∞ fndµ But also gn(x) ≤ fk(x) for each k ≥ n and so
Z
R
gndµ ≤ Z
R
fkdµ (k ≥ n)
or Z
R
gndµ ≤ inf
k≥n
Z
R
fkdµ Hence
lim inf
n→∞
Z
R
fndµ = lim
n→∞
k≥ninf Z
R
fkdµ
≥ lim
n→∞
Z
R
gndµ = Z
R
lim inf
n→∞ fndµ
Example4.1.2. Fatou’s lemma is not true with ‘equals’.
For instance take, fn = χ[n,2n] and notice thatR
Rfndµ = n → ∞ as n → ∞ but for each x ∈ R, limn→∞fn(x) = 0. So
lim inf
n→∞
Z
R
fndµ = ∞ >
Z
R
lim inf
n→∞ fndµ = Z
R
0 dµ = 0
This also shows that the Monotone Convergence Theorem is not true without ‘Monotone’.
4.2 Almost everywhere
Definition 4.2.1. We say that a property about real numbers x holds almost everywhere (with respect to Lebesgue measure µ) if the set of x where it fails to be true has µ measure 0.
Proposition 4.2.2. If f : R → [−∞, ∞] is integrable, then f (x) ∈ R holds almost everywhere (or, equivalently,|f (x)| < ∞ almost everywhere).
Proof. Let E = {x : |f (x)| = ∞}. What we want to do is show that µ(E) = 0.
We know R
R|f | dµ < ∞. So, for any n ∈ N, the simple function nχE satisfies nχE(x) ≤
|f (x)| always, and so has Z
R
nχEdµ = nµ(E) ≤ Z
R
|f | dµ < ∞.
But this can’t be true for all n ∈ N unless µ(E) = 0.
Proposition 4.2.3. If f : R → [−∞, ∞] is measurable, then f satisfies Z
R
|f | dµ = 0 if and only iff (x) = 0 almost everywhere.
Proof. SupposeR
R|f | dµ = 0 first. Let En= {x ∈ R : |f (x)| ≥ 1/n}. Then 1
nχEn ≤ |f |
and so
Z
R
1
nχEndµ = 1
nµ(En) ≤ Z
R
|f | dµ = 0.
Thus µ(En) = 0 for each n. But E1 ⊆ E2 ⊆ · · · and
∞
[
n=1
En= {x ∈ R : f (x) 6= 0}.
So
µ({x ∈ R : f (x) 6= 0}) = µ
∞
[
n=1
En
!
= lim
n→∞µ(En) = 0.
Conversely, suppose now that µ({x ∈ R : f (x) 6= 0}) = 0. We know |f | is a non-negative measurable function and so there is a monotone increasing sequence (fn)∞n=1 of measurable simple functions that converges pointwise to |f |. From 0 ≤ fn(x) ≤ |f (x)| we can see that {x ∈ R : fn(x) 6= 0} ⊆ {x ∈ R : f (x) 6= 0} and so µ({x ∈ R : fn(x) 6= 0}) ≤ µ({x ∈ R : f (x) 6= 0}) = 0. Being a simple function fnhas a largest value yn (which is finite) and so if we put En = {x ∈ R : fn(x) 6= 0} we have
fn≤ ynχEn ⇒ Z
R
fndµ ≤ Z
R
ynχEndµ = yn Z
R
χEndµ = ynµ(En) = 0.
From the Monotone Convergence Theorem Z
R
|f | dµ = Z
R
n→∞lim fn
dµ = lim
n→∞
Z
R
fndµ = lim
n→∞0 = 0.
The above result is one way of saying that integration ‘ignores’ what happens to the integrand on any chosen set of measure 0. Here is a result that says that in way that is often used.
Proposition 4.2.4. Let f : R → [−∞, ∞] be an integrable function and g : R → [−∞, ∞]
a Lebesgue measurable function with f (x) = g(x) almost everywhere. Then g must also be integrable andR
Rg dµ = R
Rf dµ.
Proof. Let E = {x ∈ R : f (x) 6= g(x)} (think of E as standing for ‘exceptional’) and note that f (x) = g(x) almost everywhere means µ(E) = 0.
Write f = (1 − χE)f + χEf . Note that both (1 − χE)f and χEf are integrable because they are measurable and satisfy |(1 − χE)f | ≤ |f | and |χEf | ≤ |f |. Also
Z
R
χEf dµ
≤ Z
R
|χEf | dµ = 0
as χEf = 0 almost everywhere. SimilarlyR
RχEg dµ = 0.
So
Z
R
f dµ = Z
R
((1 − χE)f + χEf ) dµ
= Z
R
(1 − χE)f dµ + Z
R
χEf dµ
= Z
R
(1 − χE)f dµ + 0
= Z
R
(1 − χE)g dµ
The same calculation (with |f | in place of f ) shows R
R(1 − χE)|g| dµ = R
R|f | dµ < ∞, so that (1 − χE)g must be integrable. Thus g = (1 − χE)g + χEg is also integrable (because R
R|χEg| dµ = 0 and so χEg is integrable and g is then the sum of two integrable functions).
Thus we haveR
Rg dµ = R
R(1 − χE)g dµ +R
RχEg dµ = R
Rf dµ + 0.
Remark4.2.5. It follows that we should be able to manage without allowing integrable functions to have the values ±∞. The idea is that, if f is integrable, it must be almost everywhere finite.
If we change all the values where |f (x)| = ∞ to 0 (say) we are only changing f on a set of x’s of measure zero. This is exactly changing f to the (1 − χE)f in the above proof. The changed function will be almost everywhere the same as the original f , but have finite values everywhere.
So from the point of view of the integral of f , this change is not significant.
However, it can be awkward to have to do this all the time, and it is better to allow f (x) =
±∞.
4.3 Dominated convergence theorem
Theorem 4.3.1 (Lebesgue dominated convergence theorem). Suppose fn: R → [−∞, ∞] are (Lebesgue) measurable functions such that the pointwise limit f (x) = limn→∞fn(x) exists.
Assume there is an integrableg : R → [0, ∞] with |fn(x)| ≤ g(x) for each x ∈ R. Then f is integrable as isfnfor eachn, and
n→∞lim Z
R
fndµ = Z
R
n→∞lim fndµ = Z
R
f dµ
Proof. Since |fn(x)| ≤ g(x) and g is integrable,R
R|fn| dµ ≤R
Rg dµ < ∞. So fnis integrable.
We know f is measurable (as a pointwise limit of measurable functions) and then, similarly,
|f (x)| = limn→∞|fn(x)| ≤ g(x) implies that f is integrable too.
The proof does not work properly if g(x) = ∞ for some x. We know that g(x) < ∞ almost everywhere. So we can take E = {x ∈ R : g(x) = ∞} and multiply g and each of the functions fnand f by 1 − χE to make sure all the functions have finite values. As we are changing them all only on the set E of measure 0, this change does not affect the integrals or the conclusions.
We assume then all have finite values.
Let hn= g − fn, so that hn≥ 0. By Fatou’s lemma lim inf
n→∞
Z
R
(g − fn) dµ ≥ Z
R
lim inf
n→∞ (g − fn) dµ = Z
R
(g − f ) dµ and that gives
lim inf
n→∞
Z
R
g dµ − Z
R
fndµ
= Z
R
g dµ − lim sup
n→∞
Z
R
fndµ ≥ Z
R
g dµ − Z
R
f dµ or
lim sup
n→∞
Z
R
fndµ ≤ Z
R
f dµ (1)
Repeat this Fatou’s lemma argument with g + fnrather than g − fn. We get lim inf
n→∞
Z
R
(g + fn) dµ ≥ Z
R
lim inf
n→∞ (g + fn) dµ = Z
R
(g + f ) dµ and that gives
lim inf
n→∞
Z
R
g dµ + Z
R
fndµ
= Z
R
g dµ + lim inf
n→∞
Z
R
fndµ ≥ Z
R
g dµ + Z
R
f dµ or
lim inf
n→∞
Z
R
fndµ ≥ Z
R
f dµ (2)
Combining (1) and (2) we get Z
R
f dµ ≤ lim inf
n→∞
Z
R
fndµ ≤ lim sup
n→∞
Z
R
fndµ ≤ Z
R
f dµ which forces
Z
R
f dµ = lim inf
n→∞
Z
R
fndµ = lim sup
n→∞
Z
R
fndµ
and that gives the result because if lim supn→∞an = lim infn→∞an(for a sequence (an)∞n=1), it implies that limn→∞anexists and limn→∞an= lim supn→∞an= lim infn→∞an.
Remark4.3.2. The example following Fatou’s lemma also shows that the assumption about the existence of the dominating function g can’t be dispensed with.
4.4 Applications of the dominated convergence theorem
Theorem 4.4.1 (Continuity of integrals). Assume f : R × R → R is such that x 7→ f[t](x) = f (x, t) is measurable for each t ∈ R and t 7→ f (x, t) is continuous for each x ∈ R. Assume also that there is an integrableg : R → R with |f (x, t)| ≤ g(x) for each x, t ∈ R. Then the function f[t]is integrable for eacht and the function F : R → R defined by
F (t) = Z
R
f[t]dµ = Z
R
f (x, t) dµ(x) is continuous.
Proof. Since f[t]is measurable and |f[t]| ≤ g we haveR
R|f[t]| dµ ≤R
R g dµ < ∞ and so f[t] is integrable (for each t ∈ R). This F (t) makes sense.
To show that F is continuous at t0 ∈ R it is enough to show that for each sequence (tn)∞n=1 with limn→∞tn= t0 we have limn→∞F (tn) = F (t0).
But that follows from the dominated convergence theorem applied to fn(t) = f (x, tn), since we have
n→∞lim fn(t) = lim
n→∞f (x, tn) = f (x, t0)
by continuity of t 7→ f (x, t). We also have |fn(t)| = |f (x, tn)| ≤ g(x) for each n and each x ∈ R.
Example4.4.2. Show that
F (t) = Z
[0,∞)
e−xcos(πt) dµ(x) is continuous.
Proof. The idea is to apply the theorem with dominating function g(x) given by g(x) = χ[0,∞)(x)e−x=
(e−x for x ≥ 0 0 for x < 0 We need to know thatR
Rg dµ < ∞ (and that g is measurable and that x 7→ χ[0,∞)(x)e−xcos(πt) is measurable for each t — but we do know that these are measurable because e−xis continuous and χ[0,∞) is measurable).
By the Monotone Convergence Theorem, Z
R
g dµ = lim
n→∞
Z
R
χ[−n,n]g dµ = lim
n→∞
Z
R
χ[0,n]e−xdµ(x) = lim
n→∞
Z n 0
e−xdµ(x)
You can work this out easily using ordinary Riemann integral ideas and the limit is 1. So R
Rg dµ < ∞.
Now the theorem applies because
|χ[0,∞)(x)e−xcos(πt)| ≤ g(x)
for each (x, t) ∈ R2 (and certainly t 7→ χ[0,∞)(x)e−xcos(πt) is continuous for each x).
Theorem 4.4.3 (Differentiating under the integral sign). Assume f : R × R → R is such that x 7→ f[t](x) = f (x, t) is measurable for each t ∈ R, that f[t0](x) = f (x, t0) is integrable for somet0 ∈ R and ∂f (x,t)∂t exists for each(x, t). Assume also that there is an integrable g : R → R with|∂f∂t|(x,t)| ≤ g(x) for each x, t ∈ R. Then the function x 7→ f(x, t) is integrable for each t and the functionF : R → R defined by
F (t) = Z
R
ftdµ = Z
R
f (x, t) dµ(x) is differentiable with derivative
F0(t) = d dt
Z
R
f (x, t) dµ(x) = Z
R
∂
∂tf (x, t) dµ(x).
Proof. Applying the Mean Value theorem to the function t 7→ f (x, t), for each t 6= t0 we have to have some c between t0 and t so that
f (x, t) − f (x, t0) = ∂f
∂t|(x,c)(t − t0) It follows that
|f (x, t) − f (x, t0)| ≤ g(x)|t − t0| and so
|f (x, t)| ≤ |f (x, t0)| + g(x)|t − t0|.
Thus
Z
R
|f (x, t)| dµ(x) ≤ Z
R
(|f (x, t0)| + g(x)|t − t0|) dµ(x)
= Z
R
|f (x, t0)| dµ(x) + |t − t0| Z
R
g dµ < ∞,
which establishes that the function x 7→ f (x, t) is integrable for each t.
To prove the formula for F0(t) consider any sequence (tn)∞n=1 so that limn→∞tn = t but tn 6= t for each t. We claim that
n→∞lim
F (tn) − F (t) tn− t =
Z
R
∂
∂tf (x, t) dµ(x). (3)
We have
F (tn) − F (t) tn− t =
Z
R
f (x, tn) − f (x, t)
tn− t dµ(x) = Z
R
fn(x) dµ(x) where
fn(x) = f (x, tn) − f (x, t) tn− t . Notice that, for each x we know
n→∞lim fn(x) = ∂f
∂t|(x,t)
and so (3) will follow from the dominated convergence theorem once we show that |fn(x)| ≤ g(x) for each x.
That follows from the Mean Value theorem again because there is c between t and tn(with c depending on x) so that
fn(x) = f (x, tn) − f (x, t) tn− t = ∂f
∂t|(x,c). So |fn(x)| ≤ g(x) for each x.
4.5 What’s missing?
There are quite a few topics that are very useful and that we have not covered at all. Some of the things we have covered are simplified from the way they are often stated and used.
An example in the latter category is that the Monotone Convergence Theorem and the Dom- inated Convergence Theorem are true if we only assume the hypotheses are valid almost every- where. The Monotone Convergence Theorem is still true if we assume that the sequence (fn)∞n=1 of measurable functions satisfies fn ≥ 0 almost everywhere and fn ≤ fn+1 almost everywhere (for each n). Then the pointwise limit f (x) = limn→∞fn(x) may exist only almost everywhere.
Something similar for the Dominated Convergence Theorem.
We stuck to integrals of functions f (x) defined for x ∈ R (or for x ∈ X ∈ L — which is more or less the same because we can extend them to be zero on R \ X) and we used only Lebesgue measure µ on the Lebesgue σ-algebra L . What we need abstractly is just a mea- sure space (X, Σ, λ) and Σ-measurable integrands f : X → [−∞, ∞]. By definition f is Σ- measurable if
f−1([−∞, a]) = {x ∈ X : f (x) ≤ a} ∈ Σ (∀a ∈ R),
and it follows from that condition that f−1(B) ∈ Σ for all Borel subsets B ⊆ R. We can then talk about simple functions on X (f : X → R with finite range f (X) = {y!, y2, . . . , yn}), their standard form (f = Pn
j=1yjχFj where Fj = f−1({yj})), integrals of non-negative Σ- measurable simple functions (R
Xf dλ = Pn
j=1yjλ(Fj)), integrals of non-negative measur- able f : X → [0, ∞], (generalised) Monotone convergence theorem, λ-integrable Σ-measurable f : X → [−∞, ∞], (generalised) Dominated Convergence Theorem.
To make this applicable, we would need some more examples of measures, other that just Lebesgue measure µ : L → [0, ∞]. We did touch on some examples of measures that don’t correspond so obviously to ‘total length’ as µ does. For instance if f : R → [0, ∞] is a non- negative (Lebesgue) measurable function, there is an associated measure λf: L → [0, ∞] given by λf(E) = R
Ef dµ. (We discussed λf when f was simple but the fact that λf is a measure even when f is not simple follows easily from the Monotone Convergence Theorem.) If we choose f so that R
Rf dµ = 1, then we get a probability measure from λf (and f is called a probability density function). The standard normal distribution is the name given to λf when f (x) = (1/√
2π)e−x2/2. That’s just one example.
The Radon Nikodym theorem gives a way to recognise measures λ : L → [0, ∞] that are of the form λ = λf for some non-negative (Lebesgue) measurable function f . The key thing is that µ(E) = 0 ⇒ λ(E) = 0. The full version of the Radon Nikodym theorem applies not just to measures on (R, L ) with respect to µ, but to many more general measure spaces.
And then we could have explained area measure on R2, volume measure on R3, n-dimensional volume measure on Rn, Fubini’s theorem relating integrals on product spaces to iterated inte- grals, the change of variables formula for integrals on Rn, and quite a few more topics.
One place where measure theory comes up is in defining so-called Lebesgue spaces, which are Banach spaces defined using (Lebesgue) integration. For example L1(R) is the space of inte- grable functions f : R → R with norm kf k1 =R
R|f | dµ. Or to be more precise it is the space of almost everywhere equivalence classes of such functions. (That is so that kf k1 = 0 only for the zero element of the space, a property that norms should have.) To make L1(R) complete we need
the Lebesgue integral. Hilbert spaces like L2(R) come into Fourier analysis, for instance. By definition L2(R) is the space of almost everywhere equivalence classes of measurable f : R → R that satisfyR
R|f |2dµ < ∞ with norm given by kf k2 = R
R|f |2dµ1/2
.
In short then, there is quite a range of things that the Lebesgue theory is used for (probabil- ity theory, Fourier analysis, differential equations and partial differential equations, functional analysis, stochastic processes, . . . ). My aim was to lay the basis for studying these topics later.
R. Timoney March 15, 2016
