The Lebesgue Integral of Measurable Functions | Sebastián Fernández Rivera

The definition and integrable functions

Definition (Positive part, negative part and the general integral)

Let $f : E \to \overline{\mathbb{R}}$ be a measurable function. Then

\[f = f^{+} - f^{-}\]

with

\[f^{+} = \max(0, f) \qquad \text{and} \qquad f^{-} = \max(0, -f).\]

We know that $f^{+}$ and $f^{-}$ are measurable and non-negative, so

\[\int_{E} f^{+} \, dx, \qquad \int_{E} f^{-} \, dx\]

are well defined. Therefore we may define

\[\int_{E} f \, dx = \int_{E} f^{+} \, dx - \int_{E} f^{-} \, dx\]

provided that

\[\int_{E} f^{+} \, dx < \infty \qquad \text{or} \qquad \int_{E} f^{-} \, dx < \infty.\]

In that case we say that the integral of $f$ over $E$ exists (with value possibly $\pm\infty$).

Definition (Integrable function and the space $L(E)$)

We say that $f$ is integrable, or that $f \in L(E)$, if

\[\int_{E} f^{+} \, dx < \infty \qquad \text{and} \qquad \int_{E} f^{-} \, dx < \infty.\]

Note (Integral triangle inequality and a.e. finiteness)

Since $$

= f^{+} + f^{-}$, for$f \in L(E)$$ we have that

\[\left| \int_{E} f \, dx \right| \leq \int_{E} f^{+} \, dx + \int_{E} f^{-} \, dx = \int_{E} |f| \, dx,\]

where the last equality uses linearity for non-negative functions. In particular, $f \in L(E)$ if and only if $$\int_{E}

\, dx < \infty$. Moreover, if$\int_{E}

\, dx < \infty$, then$f \in \mathbb{R}$$ a.e., by the proposition on functions with finite integral.

Monotonicity and basic properties

Theorem (Monotonicity of the general integral)

Let $f : E \to \overline{\mathbb{R}}$ and $g : E \to \overline{\mathbb{R}}$ be measurable such that

\[\int_{E} f \, dx \qquad \text{and} \qquad \int_{E} g \, dx\]

exist. If $f \leq g$ a.e. on $E$, then

\[\int_{E} f \, dx \leq \int_{E} g \, dx.\]

In particular, if $f = g$ a.e. on $E$, then

\[\int_{E} f \, dx = \int_{E} g \, dx.\]

Proof: Note that if $f \leq g$ a.e., then, pointwise where the inequality holds,

\[f^{+} = \max\{0, f\} \leq \max\{0, g\} = g^{+}, \qquad g^{-} = \max\{0, -g\} \leq \max\{0, -f\} = f^{-},\]

that is, $f^{+} \leq g^{+}$ and $g^{-} \leq f^{-}$ a.e. Hence, by the a.e. monotonicity of the integral of non-negative functions,

\[\int_{E} f^{+} \, dx \leq \int_{E} g^{+} \, dx, \qquad \int_{E} g^{-} \, dx \leq \int_{E} f^{-} \, dx,\]

and so, subtracting in the extended sense (the existence of both integrals guarantees that the form $\infty - \infty$ does not appear),

\[\int_{E} f \, dx = \int_{E} f^{+} \, dx - \int_{E} f^{-} \, dx \leq \int_{E} g^{+} \, dx - \int_{E} g^{-} \, dx = \int_{E} g \, dx.\]

If $f = g$ a.e., the two inequalities give the equality.

Theorem (Restriction of the domain, countable additivity and null sets)

Let $f : E \to \overline{\mathbb{R}}$ be measurable such that $\int_{E} f \, dx$ exists.

If $E_{1} \subseteq E$ is measurable, then $\int_{E_{1}} f \, dx$ exists.
If $E = \bigcup_{k=1}^{\infty} E_{k}$ with the $E_{k}$ measurable and pairwise disjoint, then
\[\int_{E} f \, dx = \sum_{k=1}^{\infty} \int_{E_{k}} f \, dx.\]
If $E_{1} \subseteq E$ with $m(E_{1}) = 0$, then $\int_{E_{1}} f \, dx = 0$.

Proof: For (i), note that, by monotonicity with respect to the domain for non-negative functions,

\[0 \leq \int_{E_{1}} f^{+} \, dx \leq \int_{E} f^{+} \, dx, \qquad 0 \leq \int_{E_{1}} f^{-} \, dx \leq \int_{E} f^{-} \, dx.\]

Since one of the two integrals over $E$ is finite, so is the corresponding one over $E_{1}$, and $\int_{E_{1}} f \, dx$ exists.

For (ii), we know, by additivity of the domain for non-negative functions, that

\[\int_{E} f^{+} \, dx = \sum_{k=1}^{\infty} \int_{E_{k}} f^{+} \, dx, \qquad \int_{E} f^{-} \, dx = \sum_{k=1}^{\infty} \int_{E_{k}} f^{-} \, dx.\]

Since

\[\int_{E} f^{+} \, dx = \sum_{k=1}^{\infty} \int_{E_{k}} f^{+} \, dx < \infty \qquad \text{or} \qquad \int_{E} f^{-} \, dx = \sum_{k=1}^{\infty} \int_{E_{k}} f^{-} \, dx < \infty,\]

at least one of the two series converges, and we may subtract term by term without ambiguity. Then

\[\int_{E} f \, dx = \sum_{k=1}^{\infty} \int_{E_{k}} f^{+} \, dx - \sum_{k=1}^{\infty} \int_{E_{k}} f^{-} \, dx = \sum_{k=1}^{\infty} \left( \int_{E_{k}} f^{+} \, dx - \int_{E_{k}} f^{-} \, dx \right) = \sum_{k=1}^{\infty} \int_{E_{k}} f \, dx.\]

Finally, for (iii), if $m(E_{1}) = 0$, the lemma on the integral over null sets applied to $f^{+}$ and $f^{-}$ gives

\[\int_{E_{1}} f \, dx = \int_{E_{1}} f^{+} \, dx - \int_{E_{1}} f^{-} \, dx = 0 - 0 = 0.\]

Linearity and differences of functions

Theorem (Linearity of the general integral)

Let $f, g : E \to \overline{\mathbb{R}}$ be measurable such that $\int_{E} f \, dx$ exists, and let $c \in \mathbb{R}$. Then

\[\int_{E} c f \, dx = c \int_{E} f \, dx.\]

Moreover, if $f, g \in L(E)$, then $f + g \in L(E)$ and

\[\int_{E} (f + g) \, dx = \int_{E} f \, dx + \int_{E} g \, dx.\]

Proof: Homogeneity. If $c \geq 0$, then $(cf)^{+} = c f^{+}$ and $(cf)^{-} = c f^{-}$, and the result follows from homogeneity for non-negative functions. If $c < 0$, then

\[(cf)^{+} = -c f^{-}, \qquad (cf)^{-} = -c f^{+},\]

since $cf \geq 0$ exactly where $f \leq 0$. Then

\[\int_{E} c f \, dx = \int_{E} (-c) f^{-} \, dx - \int_{E} (-c) f^{+} \, dx = -c \int_{E} f^{-} \, dx + c \int_{E} f^{+} \, dx = c \int_{E} f \, dx.\]

Additivity. If $f, g$ are integrable, then they are finite a.e., so that $f + g$ is defined a.e., and since $$

f + g

\leq

$$,

\[0 \leq \int_{E} |f + g| \, dx \leq \int_{E} |f| \, dx + \int_{E} |g| \, dx < \infty,\]

using monotonicity and linearity for non-negative functions; that is, $f + g \in L(E)$. Note that the following measurable, pairwise disjoint sets cover $E$ up to a null set (where $f$ or $g$ take infinite values):

\[\begin{aligned} &E_{1} = \{ f \geq 0,\ g \geq 0 \}, \\ &E_{2} = \{ f < 0,\ g < 0 \}, \\ &E_{3} = \{ f \geq 0,\ g < 0,\ f + g \geq 0 \}, \\ &E_{4} = \{ f < 0,\ g \geq 0,\ f + g \geq 0 \}, \\ &E_{5} = \{ f \geq 0,\ g < 0,\ f + g < 0 \}, \\ &E_{6} = \{ f < 0,\ g \geq 0,\ f + g < 0 \}. \end{aligned}\]

By additivity of the domain, it suffices to prove the additivity of the integral on each $E_{i}$.

On $E_{1}$ the three functions $f, g, f+g$ are non-negative, so, by linearity for non-negative functions,

\[\int_{E_{1}} (f + g) \, dx = \int_{E_{1}} (f^{+} + g^{+}) \, dx = \int_{E_{1}} f^{+} \, dx + \int_{E_{1}} g^{+} \, dx = \int_{E_{1}} f \, dx + \int_{E_{1}} g \, dx,\]

and likewise, on $E_{2}$, where all three are negative,

\[\begin{aligned} \int_{E_{2}} (f + g) \, dx &= \int_{E_{2}} -(f^{-} + g^{-}) \, dx = -\int_{E_{2}} (f^{-} + g^{-}) \, dx \\ &= -\int_{E_{2}} f^{-} \, dx - \int_{E_{2}} g^{-} \, dx = \int_{E_{2}} f \, dx + \int_{E_{2}} g \, dx. \end{aligned}\]

On $E_{3}$ we have $f = (f + g) + (-g)$ with $f + g \geq 0$ and $-g > 0$; hence, by linearity for non-negative functions,

\[\int_{E_{3}} f \, dx = \int_{E_{3}} \big( (f+g) - g \big) \, dx = \int_{E_{3}} (f + g) \, dx + \int_{E_{3}} (-g) \, dx = \int_{E_{3}} (f + g) \, dx - \int_{E_{3}} g \, dx,\]

where the last step uses homogeneity with $c = -1$ and that $g \in L(E_{3})$. Therefore

\[\int_{E_{3}} (f + g) \, dx = \int_{E_{3}} f \, dx + \int_{E_{3}} g \, dx.\]

Moreover, on $E_{5}$ we have $-g = -(f+g) + f$ with $-(f+g) > 0$ and $f \geq 0$, so

\[\int_{E_{5}} (-g) \, dx = \int_{E_{5}} \big( -(f+g) + f \big) \, dx = \int_{E_{5}} -(f+g) \, dx + \int_{E_{5}} f \, dx.\]

Thus, multiplying by $-1$ and rearranging, we have that

\[\int_{E_{5}} f \, dx + \int_{E_{5}} g \, dx = \int_{E_{5}} (f + g) \, dx.\]

The cases $E_{4}$ and $E_{6}$ are analogous, interchanging the roles of $f$ and $g$; finishing the rest of the proof is an exercise.

Note (Linear combinations and a warning about subtraction)

Let $f_{1}, \dots, f_{n} \in L(E)$ and $a_{1}, \dots, a_{n} \in \mathbb{R}$. Then, iterating the previous theorem, $\sum_{k=1}^{n} a_{k} f_{k} \in L(E)$ and

\[\int_{E} \left( \sum_{k=1}^{n} a_{k} f_{k} \right) \, dx = \sum_{k=1}^{n} a_{k} \int_{E} f_{k} \, dx.\]

Beware: when the integrals are not finite, it is in general not true that

\[\int_{E} (f - g) \, dx = \int_{E} f \, dx - \int_{E} g \, dx.\]

For example, consider $f = \mathbf{1}_{[n,\infty)}$ and $g = \mathbf{1}_{[n+1,\infty)}$: the difference $f - g = \mathbf{1}_{[n,n+1)}$ has integral $1$, but $\int f \, dx = \int g \, dx = \infty$ and the difference $\infty - \infty$ is not defined.

Proposition (Subtracting an integrable minorant)

Let $f$ and $\phi$ be measurable on $E$ such that $\phi \leq f$ a.e. and $\phi \in L(E)$. Then $\int_{E} f \, dx$ and $\int_{E} (f - \phi) \, dx$ exist and

\[\int_{E} (f - \phi) \, dx = \int_{E} f \, dx - \int_{E} \phi \, dx.\]

Proof: Since $\phi \leq f$ a.e., we have $-f \leq -\phi$ a.e. and so

\[f^{-} = \max(0, -f) \leq \max(0, -\phi) = \phi^{-} \quad \text{a.e.},\]

whence

\[0 \leq \int_{E} f^{-} \, dx \leq \int_{E} \phi^{-} \, dx < \infty.\]

That is, $\int_{E} f \, dx$ exists, with value in $(-\infty, \infty]$, and two cases remain according to the value of $\int_{E} f^{+} \, dx$.

If $\int_{E} f^{+} \, dx < \infty$, then $f \in L(E)$, and since also $-\phi \in L(E)$, linearity gives directly

\[\int_{E} (f - \phi) \, dx = \int_{E} f \, dx - \int_{E} \phi \, dx.\]

Now, if $\int_{E} f^{+} \, dx = \infty$, that is $f \notin L(E)$, we claim that $f - \phi \notin L(E)$. Indeed, where both are finite we have $f = \phi + (f - \phi) \leq \phi^{+} + (f - \phi)$, and since the right-hand side is non-negative,

\[f^{+} \leq \phi^{+} + (f - \phi) \quad \text{a.e.}\]

Integrating, by a.e. monotonicity and linearity for non-negative functions,

\[\infty = \int_{E} f^{+} \, dx \leq \int_{E} \phi^{+} \, dx + \int_{E} (f - \phi) \, dx,\]

and since $\int_{E} \phi^{+} \, dx < \infty$, it follows that $\int_{E} (f - \phi) \, dx = \infty$. Moreover $f - \phi \geq 0$ a.e., so

\[\int_{E} (f - \phi) \, dx = \int_{E} (f - \phi)^{+} \, dx = \infty.\]

We conclude that, in this case, both sides equal $\infty$:

\[\int_{E} (f - \phi) \, dx = \infty = \int_{E} f \, dx - \int_{E} \phi \, dx,\]

since $\int_{E} f \, dx = \infty$ and $\int_{E} \phi \, dx \in \mathbb{R}$.

Note (Products of integrable functions)

The conditions for $fg$ to be integrable are more complex than for the sum. For example, if $f \in L(E)$ and $g$ is measurable with $$

g(x)

\leq M$for$x \in E$, then$fg \in L(E)$, since$

\leq M

$and monotonicity gives$\int_{E}

\, dx \leq M \int_{E}

\, dx < \infty$$.

The convergence theorems

Theorem (Monotone convergence with an integrable minorant or majorant)

Let $\{f_{k}\}_{k=1}^{\infty}$ be a sequence of measurable functions on $E$, such that $\lim_{k \to \infty} f_{k} = f$ a.e. on $E$.

If there exists $\phi \in L(E)$ such that $\phi \leq f_{k} \leq f_{k+1}$ a.e. for every $k \geq 1$, then
\[\lim_{k \to \infty} \int_{E} f_{k} \, dx = \int_{E} f \, dx.\]
If there exists $\phi \in L(E)$ such that $f_{k+1} \leq f_{k} \leq \phi$ a.e. for every $k \geq 1$, then
\[\lim_{k \to \infty} \int_{E} f_{k} \, dx = \int_{E} f \, dx.\]

Proof: We shall prove only the increasing case; part (ii) is left as an exercise for the reader. If the first condition holds, then

\[f_{k} - \phi \geq 0 \qquad \text{and} \qquad f_{k+1} - \phi \geq f_{k} - \phi \quad \text{a.e.}\]

for $k \geq 1$, and moreover $f_{k} - \phi \to f - \phi$ a.e. Then, by the monotone convergence theorem for non-negative functions,

\[\lim_{k \to \infty} \int_{E} (f_{k} - \phi) \, dx = \int_{E} (f - \phi) \, dx.\]

By the proposition on subtracting an integrable minorant, applied to each $f_{k}$ and to $f$ (with minorant $\phi$),

\[\int_{E} (f_{k} - \phi) \, dx = \int_{E} f_{k} \, dx - \int_{E} \phi \, dx, \qquad \int_{E} (f - \phi) \, dx = \int_{E} f \, dx - \int_{E} \phi \, dx.\]

That is,

\[\lim_{k \to \infty} \int_{E} f_{k} \, dx - \int_{E} \phi \, dx = \int_{E} f \, dx - \int_{E} \phi \, dx,\]

and since $\int_{E} \phi \, dx \in \mathbb{R}$, we may cancel and conclude that $\lim_{k \to \infty} \int_{E} f_{k} \, dx = \int_{E} f \, dx$.

Example (Term-by-term integration of series of non-negative functions)

Let $f_{k} : E \to [0,\infty]$ be measurable for $k \geq 1$. Then, if we set

\[g_{m} = \sum_{k=1}^{m} f_{k},\]

we have that $0 \leq g_{m} \leq g_{m+1}$. Hence, by the monotone convergence theorem and linearity,

\[\begin{aligned} \int_{E} \left( \sum_{k=1}^{\infty} f_{k} \right) \, dx &= \int_{E} \lim_{m \to \infty} g_{m}(x) \, dx = \lim_{m \to \infty} \int_{E} g_{m}(x) \, dx \\ &= \lim_{m \to \infty} \int_{E} \sum_{k=1}^{m} f_{k}(x) \, dx = \lim_{m \to \infty} \sum_{k=1}^{m} \int_{E} f_{k}(x) \, dx = \sum_{k=1}^{\infty} \int_{E} f_{k}(x) \, dx. \end{aligned}\]

Example (A sequence converging to zero with constant integrals)

Let $E = [0,1]$ and define

\[f_{k}(x) = \begin{cases} k & \text{if } 0 \leq x \leq \frac{1}{k}, \\ 0 & \text{if } \frac{1}{k} < x \leq 1. \end{cases}\]

Then

\[\int_{E} f_{k}(x) \, dx = k \cdot m\left( \left[0, \tfrac{1}{k}\right] \right) = 1 \quad \text{for every } k,\]

but $\lim_{k \to \infty} f_{k} = 0$ a.e. on $[0,1]$: for $x > 0$, $f_{k}(x) = 0$ as soon as $k > \tfrac{1}{x}$. We ask:

When is $\displaystyle \lim_{k \to \infty} \int_{E} f_{k}(x) \, dx = \int_{E} \lim_{k \to \infty} f_{k}(x) \, dx$?

This example shows that some additional hypothesis (monotonicity, domination, etc.) is necessary.

Theorem (Fatou’s lemma)

Let $f_{k} : E \to [0,\infty]$ be a sequence of non-negative measurable functions. Then

\[\int_{E} \liminf_{k \to \infty} f_{k} \, dx \leq \liminf_{k \to \infty} \int_{E} f_{k} \, dx.\]

Proof: Recall that

\[\liminf_{k \to \infty} f_{k} = \sup_{k \geq 1} \inf_{m \geq k} f_{m} = \lim_{k \to \infty} \inf_{m \geq k} f_{m}.\]

Consider

\[g_{k} = \inf_{m \geq k} f_{m};\]

then $\{g_{k}\}$ is an increasing sequence of non-negative measurable functions with $g_{k} \to \liminf_{k \to \infty} f_{k}$. By the monotone convergence theorem,

\[\int_{E} \Big( \liminf_{k \to \infty} f_{k} \Big) \, dx = \lim_{k \to \infty} \int_{E} \inf_{m \geq k} f_{m} \, dx.\]

Note that

\[\inf_{m \geq k} f_{m} \leq f_{n} \quad \text{for } n \geq k \geq 1.\]

Then, by monotonicity, for $n \geq k$,

\[\int_{E} \inf_{m \geq k} f_{m}(x) \, dx \leq \int_{E} f_{n}(x) \, dx,\]

and taking the infimum over $n \geq k$,

\[\int_{E} \inf_{m \geq k} f_{m}(x) \, dx \leq \inf_{n \geq k} \int_{E} f_{n}(x) \, dx.\]

Now taking the limit as $k \to \infty$ on both sides,

\[\lim_{k \to \infty} \int_{E} \inf_{m \geq k} f_{m}(x) \, dx \leq \lim_{k \to \infty} \inf_{n \geq k} \int_{E} f_{n}(x) \, dx = \liminf_{k \to \infty} \int_{E} f_{k}(x) \, dx.\]

Combining with the first identity, we conclude that

\[\int_{E} \liminf_{k \to \infty} f_{k} \, dx \leq \liminf_{k \to \infty} \int_{E} f_{k} \, dx.\]

Note (Uniform bound on the integrals and the lower limit)

Note that if $f_{k} : E \to [0,\infty]$ is measurable for each $k$ and

\[\int_{E} f_{k}(x) \, dx \leq M \quad \text{for every } k,\]

then, by Fatou’s lemma,

\[\int_{E} \liminf_{k \to \infty} f_{k} \, dx \leq M.\]

Theorem (Lebesgue dominated convergence for non-negative functions)

Let $\{f_{k}\}_{k=1}^{\infty}$ be a sequence of non-negative measurable functions such that

\[\lim_{k \to \infty} f_{k} = f\]

a.e. on $E$. If there exists $\phi$ measurable with

\[0 \leq f_{k} \leq \phi \quad \text{a.e. for every } k \geq 1, \qquad \int_{E} \phi(x) \, dx < \infty,\]

then

\[\lim_{k \to \infty} \int_{E} f_{k}(x) \, dx = \int_{E} f(x) \, dx.\]

Proof: Since $f = \liminf_{k \to \infty} f_{k}$ a.e., Fatou’s lemma and the a.e. equality of the integrals give

\[\int_{E} f(x) \, dx \leq \liminf_{k \to \infty} \int_{E} f_{k} \, dx.\]

Note moreover that $0 \leq f \leq \phi$ a.e., so $f, f_{k} \in L(E)$, since their integrals are dominated by $\int_{E} \phi \, dx < \infty$.

Now consider $h_{k} = \phi - f_{k} \geq 0$ a.e. By Fatou’s lemma applied to $\{h_{k}\}$,

\[\int_{E} \liminf_{k \to \infty} (\phi - f_{k}) \, dx \leq \liminf_{k \to \infty} \int_{E} (\phi - f_{k}) \, dx.\]

Note that, since the limit of $f_{k}$ exists a.e.,

\[\liminf_{k \to \infty} (\phi - f_{k}) = \lim_{k \to \infty} (\phi - f_{k}) = \phi - f \quad \text{a.e.}\]

On the other hand, using linearity (all the functions involved are integrable), we have that

\[\liminf_{k \to \infty} \int_{E} (\phi - f_{k})(x) \, dx = \liminf_{k \to \infty} \left( \int_{E} \phi(x) \, dx - \int_{E} f_{k}(x) \, dx \right) = \int_{E} \phi(x) \, dx - \limsup_{k \to \infty} \int_{E} f_{k}(x) \, dx.\]

The last equality holds because $\liminf_{k}(-a_{k}) = -\limsup_{k} a_{k}$ for every real sequence $\{a_{k}\}$, and adding the finite constant $\int_{E} \phi \, dx$ commutes with the lower limit. Combining the three relations above and cancelling $\int_{E} \phi \, dx \in \mathbb{R}$, we conclude that

\[\limsup_{k \to \infty} \int_{E} f_{k}(x) \, dx \leq \int_{E} f(x) \, dx \leq \liminf_{k \to \infty} \int_{E} f_{k}(x) \, dx,\]

that is, the limit exists and $\lim_{k \to \infty} \int_{E} f_{k} \, dx = \int_{E} f \, dx$.

Theorem (Uniform convergence on domains of finite measure)

Let $f_{k} \in L(E)$ for $k \geq 1$. If $\lim_{k \to \infty} f_{k} = f$ uniformly on $E$ with $m(E) < \infty$, then $f \in L(E)$ and

\[\lim_{k \to \infty} \int_{E} f_{k} \, dx = \int_{E} f \, dx.\]

Proof: Exercise.

Note (The finite-measure hypothesis is essential for uniform convergence)

Note that $f_{k}(x) = \frac{1}{k}$ converges to zero uniformly on $\mathbb{R}$, but

\[\int_{\mathbb{R}} f_{k} \, dx = \infty\]

for every $k$, so the conclusion of the previous theorem fails without the hypothesis $m(E) < \infty$.