convergent sequence of simple functions for a measurable function

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.

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Extensions

Theorem

If $E \subset R$ is measurable and $f : E \to C$ is a 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 $0 \leq | ϕ_{0} | \leq | ϕ_{1} | \leq \dots \leq | f (x) |$
(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

Extended Reals (proof)

Split the function

f

into its positive part and negative part, both of which are nonnegative measurable functions. We can then apply the nonnegative statement above to each part of the function and get the sequences

{ϕ_{n}^{+}}

and

{ϕ_{n}^{-}}

. The final sequence is then given by

Complex Functions (proof)

We split the function

f

into its Real and Imaginary part and can apply the result for the extended reals to each part to get

{ϕ_{n}^{Re}}

and

{ϕ_{n}^{Im}}

. The final sequence is then given by

convergent sequence of simple functions for a measurable function

Nonnegative Real Functions

Extensions

Extended Reals (proof)

Complex Functions (proof)

References

See Also

Mentions