91Ó°ÊÓ

A step function is a piecewise constant function, meaning that it takes only a finite number of different constant values over its domain. More formally, a step function \(g: \mathbf{R} \to \mathbf{R}\) can be represented as $$ g(x) = \begin{cases} c_1, & \text{if } x \in I_1 \\ c_2, & \text{if } x \in I_2 \\ \cdots \\ c_n, & \text{if } x \in I_n \\ \end{cases} $$ where \(c_i\) are constants and \(I_i\) are non-overlapping intervals in \(\mathbf{R}\). Since a step function is a simple type of function, it is Lebesgue integrable, which means it belongs to the space \(\mathcal{L}^{p}(\mathbf{R})\) for any \(0<p<\infty\).

We will construct \(g\) using Exercise 3.47, which states: Suppose that \(0 0\), there exists a simple function \(\phi \in \mathcal{L}_c^p(\mathbf{R})\) such that \(\|f - \phi\|_p < \frac{\varepsilon}{2}\), where \(\mathcal{L}_c^p(\mathbf{R})\) denotes the set of compactly supported functions in \(\mathcal{L}^p(\mathbf{R})\). First, let's construct a simple function \(\phi\) such that \(\|f-\phi\|_{p}<\frac{\varepsilon}{2}\), by invoking the result from Exercise 3.47. Since \(\phi\) is a simple function, it can be written in the form: $$ \phi(x) = \sum_{i=1}^N a_i \chi_{E_i}(x) $$ where \(a_i\) are constants, \(N\) is a finite number, and \(\chi_{E_i}\) are characteristic functions of measurable sets \(E_i\). Now, let's partition the domain of the simple function \(\phi\) into non-overlapping intervals \(I_i\). Construct a step function \(g\) on each interval \(I_i\) such that the constant value of \(g\) on each interval is equal to the corresponding constant value of the simple function \(\phi\): $$ g(x) = \begin{cases} a_1, & \text{if } x \in I_1 \\ a_2, & \text{if } x \in I_2 \\ \cdots \\ a_N, & \text{if } x \in I_N \\ \end{cases} $$

To show that \(\|f-g\|_p < \varepsilon\), we will analyze the difference \(\|f-\phi\|_p\) and \(\|g-\phi\|_p\). First, we know \(\|f-\phi\|_{p} < \frac{\varepsilon}{2}\) by the result from Exercise 3.47. Now notice that \(g\) and \(\phi\) have the same value on each interval \(I_i\). Thus, their difference is 0 on these intervals, and their \(p\)-norm difference is also 0: $$ \|g-\phi\|_{p} = \left(\int_{\mathbf{R}} |g(x)-\phi(x)|^p dx\right)^{\frac{1}{p}} = 0. $$ By the triangle inequality for the \(p\)-norms, we have: $$ \|f-g\|_{p} \leq \|f-\phi\|_{p}+\|g-\phi\|_{p} < \frac{\varepsilon}{2} + 0 = \frac{\varepsilon}{2}. $$ Since \(\|f-g\|_{p} < \frac{\varepsilon}{2} \leq \varepsilon\), we have found a step function \(g \in \mathcal{L}^{p}(\mathbf{R})\) that meets the requirements of the problem.