The Simple Function Approximation Theorem

# The Simple Function Approximation Theorem

Recall from The Simple Function Approximation Lemma page that if $f$ is a Lebesgue measurable function defined on a Lebesgue measurable set $E$ and if $f$ is bounded on $E$ then for all $\epsilon > 0$ there exists steps functions $\varphi_{\epsilon}$ and $\psi_{\epsilon}$ such that:

**1)**$\varphi_{\epsilon}(x) \leq f(x) \leq \psi_{\epsilon}(x)$ for all $x \in E$.

**2)**$0 \leq \psi_{\epsilon}(x) - \varphi_{\epsilon}(x) < \epsilon$ for all $x \in E$.

In other words, the simple function approximation lemma tells us that for any Lebesgue measurable function $f$ on a Lebesgue measurable set $E$ that is bounded, we can find step functions which are arbitrarily close together and such that $f$ is between them.

We now prove the next result - the simple function approximation theorem.

Theorem 1 (The Simple Function Approximation Theorem): Let $f$ be an extended real-valued Lebesgue measurable function defined on a Lebesgue measurable set $E$. Then there exists a sequence $(\varphi_n(x))_{n=1}^{\infty}$ of simple functions on $E$ that converges to $f(x)$ pointwise on $E$ and such that $| \varphi_n(x) | \leq | f(x) |$ for all $x \in E$ and for all $n \in \mathbb{N}$. Furthermore, if $f(x) \geq 0$ then there exists a sequence of increasing simple functions with these properties. |

**Proof:**We first deal with the case that $f(x) \geq 0$ for all $x \in E$. For each $n \in \mathbb{N}$ let $f_n$ be the function defined for all $x \in E$ by:

\begin{align} \quad f_n(x) = \min \{ f(x), n \} \end{align}

- Since $f$ is a Lebesgue measurable function, each $f_n$ is a Lebesgue measurable function. Furthermore, $(f_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$. To show this, let $x \in E$. If $f(x)$ is finite, then for some $N \in \mathbb{N}$ we have that $f(x) < N$. So $f_N(x) = \min \{ f(x), N \} = f(x)$, and for each $n \geq N$, $f_n(x) = f(x)$, so $(f_n(x))_{n=1}^{\infty}$ converges to $f(x)$. Meanwhile, if $f(x) = \infty$ then for all $n \in \mathbb{N}$, $n < f(x)$. So $f_n(x) = \min \{ f(x), n \} = n$ and $(f_n(x))_{n=1}^{\infty}$ tends to $\infty = f(x)$.

- Now, note that for each $n \in \mathbb{N}$ we have that $f_n$ is a Lebesgue measurable function defined on a Lebesgue measurable set $E$, and $f_n$ is bounded on $E$. So for each $n \in \mathbb{N}$ let $\displaystyle{\epsilon_n = \frac{1}{n} > 0}$. Then there exists step functions $\varphi_n$ and $\psi_n$ such that $\varphi_n(x) \leq f_n(x) \leq \psi_n(x)$ for all $x \in E$ and $\displaystyle{0 \leq \psi_n(x) - \varphi_n(x) < \epsilon_n = \frac{1}{n}}$. Hence for each $n \in \mathbb{N}$ and for all $x \in E$:

\begin{align} \quad 0 \leq f_n(x) - \varphi_n(x) < \epsilon_n = \frac{1}{n} \\ \end{align}

- Therefore:

\begin{align} \quad f_n(x) - \frac{1}{n} < \varphi_n(x) < f_n(x) \end{align}

- Taking the pointwise limit of both sides of the inequality above gives us that $(\varphi_n(x))_{n=1}^{\infty}$ converges pointwise to $f(x)$ on $E$.

- Furthermore, since $f(x) \geq 0$, let $\varphi_n'(x) = \max \{ \varphi_1(x), \varphi_2(x), ..., \varphi_n(x) \}$. The maximum of a finite set of step functions is a step function, so each $\varphi_n'(x)$ is a step function. Furthermore, $\varphi_n'(x) \leq \varphi_n'(x)$ for all $x \in E$ so $(\varphi_n'(x))_{n=1}^{\infty}$ is an increasing sequence of step functions that converges pointwise to $f(x)$ on $E$.

- Now suppose that $f$ is any extended real-valued Lebesegue measurable function defined on a Lebesgue measurable set $E$. Let:

\begin{align} \quad f^+(x) = \max \{ f(x), 0 \} \quad \mathrm{and} \quad f^-(x) = \max \{ -f(x), 0 \} \end{align}

- Then $f^+(x) \geq 0$ and $f^-(x) \geq 0$ for all $x \in E$. So by the previous case there exists sequences of increasing step functions $(\varphi_n^{(1)}(x))_{n=1}^{\infty}$ and $(\varphi_n^{(2)}(x))_{n=1}^{\infty}$ that converge pointwise to $f^+(x)$ and $f^{-}(x)$ respectively. We note that $f(x) = f^+(x) - f^{-}(x)$ for all $x \in E$, so for each $n \in \mathbb{N}$ let:

\begin{align} \quad \varphi_n(x) = \varphi_n^{(1)}(x) - \varphi_n^{(2)}(x) \end{align}

- Then $(\varphi_n(x))_{n=1}^{\infty}$ is a sequence of step functions that converges pointwise to $f(x)$ on $E$ and furthermore, for all $x \in E$ and for all $n \in \mathbb{N}$:

\begin{align} \quad |\varphi_n(x)| = |\varphi_n^{(1)}(x) - \varphi_n^{(2)}(x)| = \varphi_n^{(1)}(x) - \varphi_n^{(2)}(x) \leq f^+(x) + f^-(x) = | f(x) | \quad \blacksquare \end{align}