CONVERGENCE THEOREMS AND FUBINI’S THEOREM
Terminology and Notation. Let f : E → R̄, where E ⊂ Rn is
Lebesgue measurable. It is very easy for the upper and lower Lebesgue
integrals of f to coincide: it suffices that f be measurable, and bounded
either above or below by a constant. The integral may be infinite, however.
We’ll say f has an integral if f is measurable and either the integral of f+
OR that of f− is finite.
[Fleming] calls such a function “integrable”, but we’ll reserve this term

for the case when both f+ AND
f− have finite integrals:
R
Definition. f is integrable (or f ∈ L1 (E)) if E |f |dx is finite. The L1
norm of f is;
Z
|f |L1 =
|f |dx.
E
We’ll denote by dx or dn x Lebesgue integration in the variable x; Lebesgue
measure of E ⊂ Rn (measurable) is denoted m(E) or mn (E).
Convergence Theorems.
Monotone Convergence Theorem. Let fk : Rn → R̄ be a monotone
increasing sequence of measurable functions. Assume fk ≥ 0 (or, more
generally, fk ≥ φ, φ ∈ L1 ); thus each fk has an integral. Let f be the
pointwise limit of the fk (which also has an integral.) Then:
Z
Z
f dx = lim fk dx.
k
The same result holds if (fk ) is decreasing and fk ≤ φ Ra.e., where φ ∈ L1 .
In particular, f ∈ L1 if, and only if, the limit of the fk dx (which exists)
is finite.
If the sequence (fk ) is not monotone, in general there may be a ‘mass
drop’. We only have:
Fatou’s Lemma. Suppose the sequence (fk ) of measurable functions
converges pointwise a.e. to f . If fk ≥ 0 (or bounded below by integrable)
we have:
Z
Z
f dx ≤ lim inf fk dx.
If fk ≤ φ, where φ ∈ L1 , we have:
Z
Z
f dx ≥ lim sup fk dx.
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It is easy to give examples where both inequalities are strict.
Now suppose the (fk ) satisfy |fk (x)| ≤ φ(x) for all x, where φ ∈ L1
(so each fk ∈ L1 ), and that fk (x) → f (x) pointwise a.e. Then applying
Fatou’s lemma to the sequence |fk (x) − f (x)| (bounded above by 2φ(x)),
which converges a.e. to zero, we find:
Z
lim |fk − f |dx = 0,
k
a stronger statement than convergence of the integrals. In words: dominated
pointwise convergence implies convergence in L1 , the Dominated Convergence Theorem.
Remark: The same results apply to integrals taken over a measurable
set E ⊂ Rn .
Applications.
Cavalieri’s theorem. Let f ≥ 0 be measurable. Denote by F (t) =
m{x; f (x) > t} (t ≥ 0)the ‘distribution function’ of f . Then:
Z
Z ∞
f (x)dx =
F (t)dt.
0
Thus the integral of f depends only on its distribution function.
This is easily verified for simple functions. Then use the fact that any
positive measurable function is the monotone pointwise limit of simple functions, continuity of Lebesgue measure with respect to countable increasing
union and the Monotone Convergence Theorem (twice!) Note the distribution function is monotone decreasing, hence Riemann-integrable in each
bounded set. (So we could take the integral on the right to be an improper
Riemann integral, possibly infinite.)
L1 convergence of series. Let fk : Rn → R be integrable functions.
Assume:
XZ
|fk (x)|dx < ∞.
k≥1
Then the series
in L1 norm:
P
k fk (x)
is absolutely convergent a.e., and convergence is
Z
lim
N
|f (x) −
N
X
fk (x)|dx = 0.
k=1
This is an application of dominated convergence.
2
Absolute continuity of the integral. Let f : Rn → R be integrable, f ≥ 0.
Then for all >R0 there exists δ > 0 so that if A ⊂ Rn is measurable and
m(A) < δ, then A f dx < .
Proof. This is Problem 1. Follow these two steps:
(i) For each integer k ≥ 1, define fk (x) by:
fk (x) = f (x) if |f (x)| ≤ k;
fk (x) = k if f (x) > k;
fk (x) = −k if f (x) < −k.
So |fk (x)| is bounded above by k. Use Dominated convergence to show
|fk − f |L1 → 0.
(ii) Pick N ≥ 1 so that |fN − f |L1 < , then let δ =
to show this δ satisfies:
Z
m(A) < δ ⇒
f dx < 2.
N.
Use |fN (x)| ≤ N
A
Volume under the graph. Let f : Rn → R be a function, f ≥ 0. Define
the subgraph of f as the subset of Rn+1 :
SGf = {(x, t) ∈ Rn × R; 0 ≤ t ≤ f (x)}.
If f is measurable, then SGf is a measurable subset of Rn+1 , with Lebesgue
measure given by:
Z
mn+1 (SGf ) =
f (x)dx.
Proof. First, if A ⊂ Rn is measurable and c > 0, the subgraph of cχA (x)
is A × [0, c]. (This is a measurable set in Rn+1 with measure cmn (A), see
problem 2 below.) By linearity, for simple functions:
φ=
N
X
ci χEi , ci > 0 ⇒ mn+1 (SGφ ) =
i=1
X
Z
ci mn (Ei ) =
φdx.
i
To prove the general case, consider an increasing sequence (φk ) of positive
simple functions, converging pointwise to f . By Monotone Convergence:
Z
Z
lim φk dx = f dx.
k
On the other hand, SGf = ∪k≥1 SGφk , an increasing union. So by continuity
of Lebesgue measure:
mn+1 (SGφk ) → mn+1 (SGf ).
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Remark. Conversely, if the subgraph is measurable, f is measurable.
This follows from Fubini’s theorem, since, for each t ∈ R:
{x; f (x) > t} = (SGf )t ,
the section of the subgraph at t. Fubini’s theorem implies this is a measurable subset of Rn for a.e. t, which is enough to show measurability of f
(check this.)
Problem 2. Let E ⊂ Rn be a measurable subset. Then E × [0, 1] is a
measurable subset of Rn+1 .
Hint: First assume E is bounded; use a characterization of measurable
sets.
FUBINI’S THEOREM
Fubini’s theorem characterizes measurability and the measure of subsets
of Rn+k = Rn × Rk in terms of its sections. Given E ⊂ Rn+k = Rn × Rk ,
the section of E at x ∈ Rn is the subset of Rk :
Ex = {y ∈ Rk ; (x, y) ∈ E}.
Fubini’s theorem. Let E ⊂ Rn+k be a Lebesgue measurable set. The following hold:
(i) Ex is a measurable set of Rk , for a.e. x ∈ Rn ;
(ii) The function µE (x) = mk (Ex ) is measurable in Rn (defined as 0 if
Ex is not measurable),
has an integral;
R and therefore
n
(iii) mn+k (E) = mk (Ex )d x.
In particular: mn+k (E) = 0 if, and only if, mk (Ex ) = 0 for mn -a.e.x.
Proof. 1. The claims are true if E = R×S is a rectangle (R ⊂ Rn , S ⊂ Rk
rectangles.) Then Ex = S, x ∈ R and Ex = ∅, x 6∈ R, so Ex is mk measurable for all x; and µ(x) = mk (S)χR (x), a simple function with integral mk (S)mn (R), or mn+k (E).
2. (a) The claims are true for the disjoint union E t F , if they hold for
two disjoint sets E, F . This follows easily from:
(E t F )x = Ex t Fx ,
µEtF (x) = µE (x) + µF (x).
(b) If E ⊂ F and the claims are true for E, F , then they also hold for
F \ E. This follows from (F \ E)x = Fx \ Ex , µF \E (x) = µF (x) − µE (x).
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It follows from (a) and (b) that validity of the claims is closed under
finite union (not necessarily disjoint) and intersection.
3. If the claims are true for each Ej in an increasing family of sets
(Ej )j≥1 , then they also hold for the union E = ∪j≥1 Ej .
We have (∪j≥1 E)x = ∪j≥1 Ejx , also an increasing union; and the positive measurable functions µEj (x), an increasing sequence, converge to µE (x),
which is therefore measurable. Then claim (iii) follows from Monotone Convergence and continuity of Lebesgue measure under increasing union:
Z
Z
Z
n
n
mn+k (E) = lim mn+k (Ej ) = lim mk (Ejx )d x = lim mk (Ejx )d x = mk (Ex )dn x.
j
j
j
Using 1.,2., 3. and the usual trick we see the claims in the theorem are
preserved under countable union (not necessarily increasing.)
In fact it follows from steps 1-3 that the claims are true in the following
cases (proved successively): E open, compact, closed, or a countable union
of closed sets.
Problem 3. Show that any Lebesgue measurable set A ⊂ Rn can be
expressed as the (disjoint) union of a set of measure zero and a countable
union of closed subsets of A.
Hint. Consider bounded sets first; recall the definition of inner measure
and use continuity of measure under increasing union.
Thus the only step missing is proving the claims in case E is a set of
measure zero in Rn+k .
4. If mn+k (E) = 0, we may find a decreasing sequence of open sets
(Ai )i≥1 so that:
\
E ⊂ A :=
Ai , mn+k (A1 ) < ∞, mn+k (A) = lim mn+k (Ai ) = 0.
i
i≥1
Since:
Z
mn+k A1 =
mk (A1x )dn x < ∞
we know mk (A1x ) < ∞ a,e,(x). Thus for a.e.x:
mk (Ax) = lim mk (Aix ).
i
So µA (x) is measurable, a decreasing limit of positive measurable functions.
By Monotone Convergence:
Z
Z
Z
n
n
0 = mn+k (A) = lim mn+k (Ai ) = lim mk (Aix )d x = lim mk (Aix )d x = mk (Ax )dn x.
i
i
i
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So mk (Ax ) = 0 for a.e. x. Since Ex ⊂ Ax , also mk (Ex ) = 0 a.e. Thus,
trivially:
Z
mk (Ex )dn x = 0 = mn+k (E).
Iterated integration: Tonelli’s theorem.
Theorem. Let f : Rn+k → R have an integral over a measurable set E;
for x ∈ Rn , denote by Ex the slice of E at x (which is measurable in Rk for
a.e. x.) Then:
(i) The function φx (y) = f (x, y) is measurable on Ex and has an integral,
k
for a.e. x. (Here (x, y) ∈ Rn ×
R R .)
(ii) The function Φ(x) = Ex φx (y)dk y is measurable in Rn , and has an
integral.R
R
(iii) E f (z)dn+k z = Φ(x)dn x, or equivalently:
Z
Z Z
f (z)dn+k z = (
f (x, y)dk y)dn x.
E
Ex
Proof. Assume first f ≥ 0. Denote by SGf,E ⊂ Rn+k+1 the subgraph of
f over E, a measurable set. Then for each x ∈ Rn :
SGφx ,Ex = {(y, t) ∈ Rk+1 ; 0 ≤ t ≤ φx (y) = f (x, y)} = (SGf,E )x ,
the section at x of the subgraph of f . By Fubini’s theorem, this set is
measurable in Rk+1 for a.e. x; for any such x, φx is measurable (and hence
has an integral). This proves (i)
R
As seen earlier, Φ(x) = Ex φx (y)dk y = mk+1 (SGφx ,Ex ) = mk+1 [(SGf,E )x ].
By Fubini’s theorem, this is a measurable function of x, hence has an integral
(given positivity.)This proves (ii)
To prove (iii), consider:
Z
n+k
f (z)d
Z
z = mn+k+1 (SGf,E ) =
n
mk+1 [(SGf,E ))x ]d x =
Z
Φ(x)dn x,
by Fubini’s theorem.
To deal with the general case, it is enough to consider the decomposition
f = f+ − f− .
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Preservation of measurable sets by locally Lipschitz maps.
Proposition. Let f : Rn → Rn be a locally Lipschitz map. Then f maps
measurable sets to measurable sets, and null sets to null sets.
Proof. f maps compact sets to compact sets, and any closed set is
a countable increasing union of compact sets. Hence f (E) is a countable
increasing union of compact sets (and therefore measurable) if E is closed.
By problem 3 above, it is enough to show:
Problem 4. f (locally Lipschitz) maps sets of outer measure zero to
sets of outer measure zero.
Hint. Let K ⊂ Rn be compact, LK a Lipschitz constant for f in K. Show
that f maps any cube Qr ⊂ K of side length r to a set f (Qr ) contained in
√
a cube of side length r nLK . Explain why this implies:
m∗ (f (E ∩ K)) ≤ nn/2 LnK m∗ (E ∩ K)
and conclude the proof.
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