Real Analysis, Folland problem 2.14. Integration of Nonnegative functions

This comes from Real Analysis, by Folland. We will use proposition 2.15 in part of the proof which is:

Proposition 2.15 - If $\{f_n\}$ is a finite or infinite sequence in $L^{+}$ and $f = \sum_{n}f_n$, then $\int f = \sum_{n}\int f_n$

Problem 2.14 - If $f\in L^{+}$, let $\lambda(E) = \int_{E}f d\mu$ for $E\in M$. Then $\lambda$ is a measure on $M$, and for any $g\in L^{+}$, $\int g d\lambda = \int f g d\mu$.(First suppose that $g$ is simple)

Proof: Observe that $\lambda(\emptyset) = \int_{\emptyset}f d\mu = \int 1_{\emptyset} f d\mu = \int 0 f d\mu = 0$. Let $\{E_n\}_{n\in\mathbb{N}}\subset M$ and let $F = \bigcup_{n=1}^{\infty}E_n$. Then $$\lambda(F) = \int_{F}f d\mu = \int 1_{F}f d\mu = \int \left(\sum_{n=1}^{\infty}1_{E_n}f\right)d\mu = \sum_{n=1}^{\infty}\int 1_{E_n}f d\mu \ \ \text{by proposition 2.15}\\ = \sum_{n=1}^{\infty}\int_{E_n}f d\mu = \sum_{n=1}^{\infty}\lambda(E_n)$$ Therefore $\lambda$ is a measure. Now, let $g\in L^{+}$, where $g$ is a simple with standard representation $g = \sum_{n=1}^{N}a_n 1_{E_n}$, then $$\int g d\lambda = \sum_{n=1}^{N}a_n\lambda(E_n) = \sum_{n=1}^{N}a_n\int_{E_n}f d\mu = \sum_{n=1}^{N}a_n\int f 1_{E_n}d\mu$$ $$=\int \sum_{n=1}^{N}a_n f 1_{E_n}d\mu = \int f g d\mu$$ Otherwise, there exists an increasing sequence $\{g_n \}_{n\in\mathbb{N}}\in L^{+}$ that converges to $g$, so that $\{fg_n\}_{n\in\mathbb{N}}$ converges to $fg$ and hence $$\int g d\lambda = \lim_{n\rightarrow \infty}\int g_n d\lambda = \lim_{n\rightarrow \infty}\int f g_n d\mu = \int f g d\mu$$

I just want to check if this approach is correct, any suggestions is greatly appreciated.

• You're not done yet - you need to do it for any $g\ge0$, not just simple $g$. Commented Nov 10, 2015 at 18:58
• See edit version I think that takes care of that Commented Nov 10, 2015 at 19:43
• Yes, except you didn't say what you meant at one point Commented Nov 10, 2015 at 22:55

Clearly $\lambda(E) \ge 0$ for all $E \in \mathcal{M}.$ Moreover, $\lambda(\varnothing) = \int_\varnothing f \,d\mu = \int f \chi_\varnothing \,d\mu = \int 0 \,d\mu = 0.$ If $\{ E_n \}_{n \in \mathbb{N}}$ is a pairwise disjoint subcollection of $\mathcal{M}$ then $(f \chi_{\bigcup\limits_{n = 1}^N E_n})_{N \in \mathbb{N}}$ is a sequence of measurable functions increasing to $f \chi_{\bigcup\limits_{n \in \mathbb{N}} E_n}$, so \begin{align*} \lambda(\bigcup\limits_{n \in \mathbb{N}} E_n) &= \int_{\bigcup\limits_{n \in \mathbb{N}} E_n} f \,d\mu \\ &= \int f \chi_{\bigcup\limits_{n \in \mathbb{N}} E_n} \,d\mu \\ &= \lim\limits_{N \to \infty} \int f \chi_{\bigcup\limits_{n = 1}^N E_n} \,d\mu \\ &= \lim\limits_{N \to \infty} \int f \sum\limits_{n = 1}^N \chi_{E_n} \,d\mu \\ &= \lim\limits_{N \to \infty} \sum\limits_{n = 1}^N \int f \chi_{E_n} \,d\mu \\ &= \lim\limits_{N \to \infty} \sum\limits_{n = 1}^N \int_{E_n} f \,d\mu \\ &= \sum\limits_{n = 1}^\infty \int_{E_n} f \,d\mu \\ &= \sum\limits_{n = 1}^\infty \lambda(E_n) \end{align*} by the monotone convergence theorem. Therefore $\lambda$ is a measure. Now let $g \in L^+.$ If $g$ is simple with standard representation $\sum\limits_{n = 1}^N a_n \chi_{E_n},$ then $$\int g \,d\lambda = \sum\limits_{n = 1}^N a_n \lambda(E_n) = \sum\limits_{n = 1}^N a_n \int_{E_n} f \,d\mu = \sum\limits_{n = 1}^N a_n \int f \chi_{E_n} \,d\mu = \int \sum\limits_{n = 1}^N a_n f \chi_{E_n} \,d\mu = \int f g \,d\mu.$$ Otherwise, there exists an increasing sequence $(g_n)_{n \in \mathbb{N}}$ of simple functions in $L^+$ which converges pointwise to $g,$ so that $(f g_n)_{n \in \mathbb{N}}$ increases pointwise to $f g$ and hence $$\int g \,d\lambda = \lim\limits_{n \to \infty} \int g_n \,d\lambda = \lim\limits_{n \to \infty} \int f g_n \,d\mu = \int f g \,d\mu,$$ by two applications of the monotone convergence theorem.