Question 11.T.1: Let f ∈ L+^0(Ω) and suppose (φn) is a sequence in S+(Ω) such...
Let f ∈ \mathcal{L}_{+}^{0} (Ω) and suppose (\varphi_{n}) is a sequence in \mathcal{S}_{+}(Ω) such that \varphi_{n} \nearrow f. Then
\underset{n→∞}{\lim} \int_{Ω}{\varphi_{n}} dm = \int_{Ω}{f} dm.
The monotonicity property of the integral over \mathcal{S}_{+} guarantees the existence of the limit. By the definition of \int_{Ω}f dm we have
\underset{n→∞}{\lim} \int_{Ω}{\varphi_{n}} dm ≤ \int_{Ω}{f} dm,
and we have to prove this inequality in the reverse direction. Suppose φ is any function in \mathcal{S}_{+}(Ω) such that φ ≤ f. The sequence (\varphi ∧\varphi_{n}) is then monotonically increasing in \mathcal{S}_{+}(Ω) and converges to φ. Using Lemma 11.2, we conclude that
\int_{Ω}{\varphi} dm = \underset{n→∞}{\lim} \int_{Ω}{(\varphi ∧ \varphi_{n})}dm.
Since \varphi ∧ \varphi_{n} ≤ \varphi_{n},
\int_{Ω}{(\varphi ∧ \varphi_{n})} dm ≤ \int_{Ω}{\varphi_{n}} dm
⇒ \int_{Ω}{\varphi} dm ≤ \underset{n→∞}{\lim} \int_{Ω}{\varphi_{n}} dm. (11.9)
But φ is an arbitrary function in \mathcal{S}_{+}(Ω) which satisfies φ ≤ f , hence, by Definition 11.1 and inequality (11.9), we have
\int_{Ω}{f} dm = \sup \left\{\int_{Ω}{\varphi} dm : \varphi ∈ \mathcal{S}_{+}(Ω), \varphi ≤ f \right\} ≤\underset{n→∞}{\lim} \int_{Ω}{\varphi_{n}} dm..