Question 10.T.20: If the function f : Ω → [0, ∞] is Lebesgue measurable, then ...
If the function f : Ω → [0, ∞] is Lebesgue measurable, then there is a sequence (\varphi_{n} : n ∈ \mathbb{N}) of non-negative simple functions such that
(i) \varphi_{n+1}(x) ≥ \varphi_{n}(x) for all n ∈ \mathbb{N} and x ∈ Ω,
(ii)\underset{n→∞}{\lim}\varphi_{n}(x) = f (x) for all x ∈ Ω.
We shall often abbreviate (i) and (ii) by writing \varphi_{n} \nearrow f.
For every n ∈ \mathbb{N} we define the sets
F_{n,i} = f^{−1}([(i − 1)/2^{n}, i/2^{n})), i ∈ \left\{1, 2, · · · , n2^{n}\right\},
F_{n,∞} = f^{−1}([n, ∞]) = f^{−1}([n, ∞) ∪ f^{−1}(\left\{∞\right\}),
and the simple function
\varphi_{n} = \sum\limits_{i=1}^{n2^{n}}{\frac{i − 1}{2^{n}}} χ_{F_{n,i}} + nχ_{F_{n,∞}} ,
which is measurable since all of the intervals \left[\frac{i − 1}{2^{n}},\frac{i}{2^{n}} \right] and [n, ∞) are Borel sets.
(i) Let x ∈ F_{n,i} where 1≤ i ≤ n2^{n}, that is
\frac{i − 1}{2^{n}} ≤ f (x) < \frac{i}{2^{n}}.
There are two possibilities:
(a) \frac{i − 1}{2^{n}} ≤ f (x) < \frac{2i − 1}{2^{n+1}},
(b) \frac{2i − 1}{2^{n+1}} ≤ f (x) < \frac{i}{2^{n}}.
In case (a), x ∈ F_{n+1,2i−1} and therefore
\varphi_{n+1}(x) = \frac{2i − 2}{2^{n+1}} = \frac{i − 1}{2^{n}} = \varphi_{n}(x).
In case (b), x ∈ F_{n+1,2i} and therefore
\varphi_{n+1}(x) = \frac{2i − 1}{2^{n+1}} > \frac{i − 1}{2^{n}} = \varphi_{n}(x).
If x ∈ F_{n,∞} then either n ≤ f(x) < n + 1, or f(x) ≥ n + 1. In the first case,
\frac{i − 1}{2^{n+1}} ≤ f (x) < \frac{i }{2^{n+1}}, n2^{n+1} +1 ≤ i < (n + 1)^{2n+1}
⇒ \varphi_{n+1}(x) = \frac{i − 1}{2^{n+1}} ≥ n = \varphi_{n}(x).
In the second case, \varphi_{n+1}(x) = n + 1 > n = \varphi_{n}(x).
(ii) If f(x) = ∞, then \varphi_{n}(x) = n → ∞. Suppose, therefore, that f(x) < N for some positive integer N. For all n ≥ N , there is a positive integer i ≤ 2^{n}N such that
\frac{i − 1}{2^{n}} ≤ f (x) < \frac{i}{2^{n}},
which implies
0 ≤ f (x) − \frac{i − 1}{2^{n}} = f (x) − \varphi_{n}(x) < \frac{1}{2^{n}}.
Hence \varphi_{n}(x) → f (x).
