Exercise 1.3.24 (Approximation of measurable functions by continuous functions)

This is a baby version of Lusin’s theorem.

• We prove this assertion by expanding to an always more general case.

  • $f$ is an indicator function $1_E$ for some Lebesgue measurable set $E\subseteq {ℝ}^d$
    We define a sequence of functions ${\left(g\right)}_{n\in \mathbb{N}}$ given by:

    $$g_n:{ℝ}^d\to [0,1]g(x)=\max <!--\; FUNCTION\; APPLICATION\; -->\{1-n\cdot \mathrm{dist}(x,E),0\}.$$

    Then the sequence ${\left(g\right)}_{n\in \mathbb{N}}$ converges to $f$ pointwise. Whenever $x\notin E$ we can find $n\in \mathbb{N}$ large enough so that $1-n\cdot \mathrm{dist}(x,E)$ becomes negative, whereas for $x\in E$ this quantity will always remain 1.

  • $f$ is a simple function
    By definition we can write $f$ as a linear combination of indicator functions $c_1\cdot 1_{E_1}+\dots +c_k{1}_{E_k}$ for $c_1,\dots ,c_k\in ℝ$ and $E_1,\dots ,E_k\subseteq {ℝ}^d$ disjoint Lebesgue measurable. Since every $1_{E_i}$ is a limit of continuous functions, by the limit laws $f$ is the limit of the continuous functions as well.
  • $f$ is Lebesgue measurable
    By definition there exists a sequence ${(s_n)}_{n\in \mathbb{N}}$ of simple functions which converges pointwise to $f$. Since each $s_n$ is a limit of continuous functions, using countable axiom of choice we can easily construct a sequence of continuous functions converging to $f$.
• Suppose that $f$ is the (almost everywhere) pointwise limit of continuous functions ${\left(f\right)}_{n\in \mathbb{N}}$. By part (i) of Exercise 1.3.8 every continuous function $f_n$ is Lebesgue measurable, and by part (iv) of the same exercise an (almost everywhere) pointwise limit $f$ of the Lebesgue measurable functions ${\left(f\right)}_{n\in \mathbb{N}}$ is again Lebesgue measurable.