Density of simple functions in Lp

Remarks

For $f \ge 0$ in $L^p$, take simple functions $0 \le s_n \le f$ that increase pointwise to $f$. Each $s_n$ vanishes outside a set of finite measure, because $s_n^p \le f^p$ is integrable. Since $\abs{f - s_n}^p \le f^p$, the dominated convergence theorem gives $\norm{f - s_n}_p \to 0$. A general $f$ is split into the positive and negative parts of its real and imaginary parts.

The hypothesis $p < \infty$ is necessary. On $\R$ with Lebesgue measure, the constant function $1$ lies in $L^\infty$, but every $s \in S$ vanishes outside a set of finite measure, so $\norm{1 - s}_\infty \ge 1$. In $L^\infty(\mu)$ the set of all simple functions, without the condition $\mu(A_i) < \infty$, is dense.

On $\R^n$ the result is the first step towards the density of step functions and of continuous functions with compact support in $L^p$, $1 \le p < \infty$.

Going deeper

For a student

The standard way to prove an inequality or an identity on $L^p$ is to check it on a dense class and extend it by continuity. Simple functions are the first such class; they are used in the proof of the dual of Lp and in interpolation theorems such as the Riesz–Thorin theorem. For $p = \infty$ the density of all simple functions is a uniform approximation, which is why the finite measure condition must then be dropped.