Functional Analysis Lecture 10

[[lecture-data]]

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Lecture Notes: Rodriguez, page 47

2. Lebesgue Measure and Integration

Recall

Measurable function

Let $E \subset R$ be measurable and $f : E \to [- \infty, + \infty]$ . We say $f$ is a Lebesgue measurable function is for all $α \in R$ .
$f^{- 1} ((α, \infty]) \in M$
ie, the preimage is a measurable set.

We also showed:

All of these had to do with functions on the extended reals, but we might also deal with complex-valued functions. Thus we will extend our existing definition of a measurable function:

Measurable Function (extension to complex image)

If $E \subset R$ is measurable, a complex-valued function $f : E \to C$ is measurable if both

$Re (f) : E \to R$ and
$Im (f) : E \to R$
are measurable.

We can also verify that the results from last lecture also hold:

Theorem

If $f, g : E \to C$ are measurable and $α \in C$ , then

$α f$
$f + g$
$f g$
$\bar{f}$
$| f |$

Are all measurable as well.

Proof

Exercise

see sums and products of measurable functions are measurable

Theorem

If $f_{n} : E \to C$ is measurable for all $n$ and $f_{n} (x) \to f (x)$ pointwise for all $x \in E$ , then $f$ is measurable.

Proof

Note that

lim_{n \to \infty} f_{n} (x) = f (x) ⟺ lim_{n \to \infty} Re (f_{n} (x)) = Re (f (x)) and lim_{n \to \infty} Im (f_{n} (x)) = Im (f (x))

And since the limit of a convergent sequence of measurable functions is measurable for real-valued functions, we preserve measurability for both $Re (f (x))$ and $Im (f (x))$ . sums and products of measurable functions are measurable gives us the desired result

This extends the limit of a convergent sequence of measurable functions is measurable

Note

Since we can write complex valued functions as a sum of $Re$ and $Im$ , it is not too hard to extend the results we have for real-valued functions.

Simple Function

A measurable function $ϕ : E \to C$ is simple if $| ϕ (E) | < \infty$
(ie, the size of the range is finite)

see simple function

Remark

simple functions can be written as a complex linear combination of finitely many indicator functions.

Proof

Suppose $ϕ : E \to C$ is a simple function where the range $ϕ (E) = {a_{1}, a_{2}, \dots, a_{n}}$ . Define the sets

A_{i} = ϕ^{- 1} ({a_{i}})

Note that each $A_{i}$ is measurable because they are intersections of the measurable sets where $Re (ϕ) = Re (a_{i})$ and $Im (ϕ) = Im (a_{i})$ .

For all $i \neq j$ we have $A_{i} \cap A_{j} = \emptyset$ and $⋃_{i = 1}^{n} A_{i} = E$ . ie, the $A_{i}$ form a finite partition of the domain $E$ . Thus, for all $x \in E$ , we can write

ϕ (x) = \sum_{i = 1}^{n} a_{i} \cdot 1_{A_{i}} (x)

Which is a finite complex linear combination of indicator functions.

\begin{matrix} ◼ \end{matrix}

see simple functions can be written as a finite complex linear combination of indicator functions

Recall that indicator functions are measurable and simple functions can be written as a finite complex linear combination of indicator functions. Thus the the complex extension of sums and products of measurable functions are measurable gives us that simple functions are measurable as well.

see simple functions are measurable

Note

The idea here is that every measurable set is approximately a simple function

Theorem

Scalar multiples, linear combinations, and products of simple functions are also simple functions.

Proof

This can be seen by checking that the resulting functions are measurable and that their ranges are finite.

see sums and products of simple functions are simple functions

Theorem

If $f : E \to [0, \infty]$ is a nonnegative measurable function, then there exists a sequence of simple functions ${ϕ_{n}}$ such that

(pointwise increasing sequence dominated by $f$ ) For all $x \in E$ we have $0 \leq ϕ_{0} (x) \leq ϕ_{1} (x) \leq \dots \leq f (x)$
(pointwise convergence) for all $x \in E$ we have $lim_{n \to \infty} ϕ_{n} (x) = f (x)$
(uniform convergence when $f$ is bounded) for all $B \geq 0$ , $ϕ_{n} \to f$ uniformly on the set ${x \in E : f (x) \leq B}$ where the bound holds

We start with the reals, but the proof will extend easily to the extended reals and complex numbers.

Proof

We will build our functions $ϕ_{n}$ to have better resolution and larger range as a function of $n$ .

Example

$ϕ_{0}$ will only tell whether the target function $f \geq 1$

ϕ_{0} (x) = {\begin{cases} 0 & f (x) \leq 1 \\ 1 & 1 < f (x) \end{cases}

$ϕ_{1}$ will only go up to 2 and have resolution of $\frac{1}{2}$

ϕ_{1} (x) = {\begin{cases} 0 & f (x) \leq \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} < f (x) \leq 1 \\ 1 & 1 < f (x) \leq \frac{3}{2} \\ \frac{3}{2} & \frac{3}{2} < f (x) \leq 2 \\ 2 & 2 < f (x) \end{cases}

For each $n \geq 0$ , define the sets

\begin{aligned} E_{k}^{(n)} & = {x \in E : k 2^{- n} \leq f (x) < (k + 1) 2^{- n}} \\ = f^{- 1} ((k 2^{- n}, (k + 1) 2^{- n}]) \end{aligned}

For each $0 \leq k \leq 2^{2 n} - 1$ . This yields an "interval of length $2^{- n}$ " in the range. These are measurable (inverse image of measurable functions of all borel sets are measurable), so we can define the sets

F^{(n)} = f^{- 1} ((2^{n}, \infty])

So we have that $E = F^{(n)} \cup (⋃_{k = 0}^{2^{2 n} - 1} E_{k}^{(n)})$ for all $n$ . Then, we can write

ϕ_{n} = 2^{n} 1_{F^{(n)}} + \sum_{k = 0}^{2^{2 n} - 1} \frac{k}{2^{n}} 1_{E_{k}^{(n)}} = 2^{n} 1_{F^{(n)}} + \sum_{k = 1}^{2^{2 n} - 1} \frac{k}{2^{n}} 1_{E_{k}^{(n)}}

Example

$ϕ_{1} = \frac{1}{2} \cdot 1_{f^{- 1} (\frac{1}{2}, 1]} + 1 \cdot 1_{f^{- 1} (1, \frac{3}{2}]} + \frac{3}{2} 1_{f^{- 1} (\frac{3}{2}, 2]} + 2 \cdot 1_{f^{- 1} (2, \infty]}$

Claim: this sequence of approximations satisfies the three conditions we want.

It is easy to see that $ϕ_{n}$ takes on finitely many values for each $n$ ( $2^{n} + 1$ , to be exact), so each $ϕ_{n}$ is indeed a simple function
By design, we always have $0 \leq ϕ_{n} \leq f$ since we defined each function such that $\frac{k}{2^{n}} < f (x) \leq \frac{k + 1}{2^{n}} ⟹ ϕ_{n} (x) = \frac{k}{2^{n}} < f (x)$

Pointwise Increasing

To show that the ${ϕ_{n}}$ are pointwise increasing, we note that if $x \in E_{k}^{(n)}$ , we have

\begin{aligned} \frac{k}{2^{n}} < f (x) \leq \frac{k + 1}{2^{n}} & ⟹ \frac{2 k}{2^{n + 1}} < f (x) \leq \frac{2 (k + 1)}{2^{n + 1}} \\ ⟹ x \in E_{2 k}^{(n + 1)} \cup E_{2 k + 1}^{(n + 1)} \\ ⟹ ϕ_{n} (x) = {\begin{cases} \frac{k}{2^{n}} = \frac{2 k}{2^{n + 1}} = ϕ_{n + 1} (x), & x \in E_{2 k}^{(n + 1)} \\ \frac{k}{2^{n}} = \frac{2 k}{2^{(n + 1)}} < \frac{2 (k + 1)}{2^{n + 1}} = ϕ_{n + 1} (x), & x \in E_{2 k + 1}^{(n + 1)} \end{cases} \end{aligned}

ie, we have $ϕ_{n} (x) \leq ϕ_{n + 1} (x)$ if $x \in E_{k}^{(n)}$ . If $x \in F^{(n)}$ , then we get a similar result with the same argument.

Pointwise convergence + Uniform convergence on sets where $f$ is bounded by $B$

Claim: For all $x \in {y \in E : f (y) \leq 2^{n}}$ , we have

0 \leq f (x) - ϕ_{n} (x) \leq 2^{- n}

Recall that each $ϕ_{n}$ partitions the range into intervals of length $\frac{1}{2^{n}}$ and note that

{y \in E : f (x) \leq 2^{n}} = ⋃_{k = 0}^{2^{2 n} - 1} E_{k}^{(n)}

Thus, we can just verify the claim for each $E_{k}^{(n)}$ . So suppose $x \in E_{k}^{(n)}$ . Then

\begin{aligned} ϕ_{n} (x) & = \frac{k}{2^{n}} \leq f (x) < \frac{k + 1}{2^{n}} \\ ⟹ f (x) - ϕ_{n} (x) & \leq \frac{k + 1}{2^{n}} - \frac{k}{2^{n}} \\ = \frac{1}{2^{n}} \end{aligned}

Pointwise convergence

Let $x \in E$ . If $f (x) = \infty$ , then we are done. Otherwise, $f^{- 1} (x) \in {y \in E : f (y) \leq 2^{n}}$ for all $n \geq N$ for some $N$ large enough. But then for all $n \geq N$ , we have

| f (x) - ϕ_{n} (x) | \leq \frac{1}{2^{n}}

ie $ϕ_{n} (x) \to f (x)$ for all $x$ .

Uniform convergence for a fixed bound

Now, for any fixed $B$ , pick some $N$ such that ${x \in E : f (x) \leq B} \subset {x \in E : f (x) \leq 2^{N}}$ . Then in the bound, we have uniform convergence.

\begin{matrix} ◼ \end{matrix}

see convergent sequence of simple functions for a measurable function

Positive and Negative part of a function

Let $f : E \to [- \infty, \infty]$ be a measurable function. Then the positive part and the negative part of $f$ is

f^{+} (x) = max {f (x), 0} f^{-} (x) = max {- f (x), 0}

respectively.

Note

$f (x) = f^{+} (x) - f^{-} (x)$
$| f (x) | = f^{+} (x) + f^{-} (x)$

See positive and negative part of a function

Now we can extend our convergent sequence of simple functions for a measurable function to the extended reals and the complex numbers.

Lebesgue Integrals

We can now define an integral by looking at the limit of a convergent sequence of simple functions for a measurable function. We don't want to depend on the simple function representation, so instead we will define something else.

First, we will build a notion of integral for nonnegative functions and then generalize to complex-valued functions from there.

L^{+}

functions

If $E \subset R$ is measurable, the class of nonnegative measurable functions on $E$ is given as

L^{+} (E) = {f : E \to [0, \infty] | f is measurable}

see nonnegative measurable functions

Lebesgue integral (for nonnegative measurable functions)

Let $ϕ \in L^{+} (E)$ be a simple, nonnegative measurable function such that $ϕ = \sum_{i = 1}^{n} a_{j} 1_{A_{i}}$ where $A_{i} \cap A_{j} = \emptyset$ for all $i, j$ and $⋃_{i = 1}^{n} A_{i} = E$ . Then the Lebesgue integral of $ϕ$ is

\int_{E} ϕ = \sum_{i = 1}^{n} a_{i} m (A_{i}) \in [0, \infty]

see Lebesgue Integral

Theorem

Suppose $ϕ, ψ$ are two simple functions. Then for any $c \geq 0$ , we have

$\int_{E} c ϕ = c \int_{E} ϕ$
$\int_{E} (ϕ + ψ) = \int_{E} ϕ + \int_{E} ψ$
$\int_{E} ϕ \leq \int_{E} ψ$ if $ϕ < ψ$
If $F \subset E$ is measurable, then $\int_{F} ϕ = \int_{E} 1_{F} ϕ \leq \int_{E} ϕ$

Proof

(1)

Note that if $ϕ = \sum_{i = 1}^{n} a_{i} 1_{A_{i}}$ then $c ϕ = \sum_{i = 1}^{n} (c a_{i}) 1_{A_{i}}$ . Thus

\begin{aligned} \int_{E} c ϕ & = \sum_{i = 1}^{n} c a_{i} m (A_{i}) \\ = c \sum_{i = 1}^{n} a_{i} m (A_{i}) \\ = c \int_{E} ϕ \end{aligned}

(2)

We again write $ϕ = \sum_{i = 1}^{n} a_{i} 1_{A_{i}}$ and $ψ = \sum_{j = 1}^{ℓ} b_{j} 1_{B_{j}}$ . Then

\begin{aligned} E & = ⋃_{i = 1}^{n} A_{i} = ⋃_{j = 1}^{ℓ} B_{j} \\ ⟹ A_{i} & = ⋃_{j = 1}^{ℓ} (A_{i} \cap B_{j}), B_{j} = ⋃_{i = 1}^{n} (A_{i} \cap B_{j}) \end{aligned}

And each set in each of the unions are disjoint since the $A_{i}$ are pairwise disjoint and the $B_{j}$ are pairwise disjoint too. Then from measure satisfies countable additivity, we get

\begin{aligned} \int_{E} ϕ + \int_{E} ψ & = \sum_{i = 1}^{n} a_{i} m (A_{i}) + \sum_{j = 1}^{ℓ} b_{j} m (B_{j}) \\ = \sum_{i = 1}^{n} \sum_{j = 1}^{ℓ} a_{i} m (A_{i} \cap B_{j}) + \sum_{j = 1}^{ℓ} \sum_{i = 1}^{m} b_{j} (A_{i} \cap B_{j}) \\ = \sum_{i, j} (a_{i} + b_{j}) m (A_{i} \cap B_{j}) \\ (*) & = \int_{E} ϕ + ψ \end{aligned}

Where $(*)$ is because we can write $ϕ + ψ = \sum_{i, j} (a_{i} + b_{j}) 1_{A_{i} \cap B_{j}}$ .

Note that this is not technically the "canonical" decomposition for simple functions since it is possible for $a_{i} + b_{j}$ to be equal to each other for different $(i, j)$ pairs. This is OK though, because we can just combine the disjoint sets where they are equal.

(3)

Write $ϕ = \sum_{i = 1}^{n} a_{i} 1_{A_{i}}$ and $ψ = \sum_{j = 1}^{ℓ} b_{j} 1_{B_{j}}$ . Then $ϕ (x) \leq ψ (x) ⟺ a_{i} \leq b_{j}$ whenever $A_{i} \cap B_{j} \neq \emptyset$ . So since we have measure satisfies countable additivity, we get

\begin{aligned} \int_{E} ϕ & = \sum_{i = 1}^{n} a_{i} m (A_{i}) \\ = \sum_{i = 1}^{n} a_{i} \sum_{j = 1}^{ℓ} m (A_{i} \cap B_{j}) \\ (A_{i} \cap B_{j} \neq \emptyset ⟹ a_{i} \leq b_{j}) ⟹ & \leq \sum_{i = 1}^{n} b_{j} \sum_{j = 1}^{ℓ} m (A_{i} \cap B_{j}) \\ = \sum_{j = 1}^{ℓ} b_{j} m (B_{j}) \\ = \int_{E} ψ \end{aligned}

(4)

Exercise

\begin{matrix} ◼ \end{matrix}

Defined the "under the curve" notion for our Lebesgue Integral

Reimann integrable functions will be Lebesgue integrable since simple functions are Riemann integrable

Next time, we will define the integral of a nonnegative measurable function

Created 2025-07-01 Last Modified 2025-07-14