I am trying to prove the follow exercise without using the Monotone Convergence Theorem.
Let $(X, \mathfrak{M}, \mu)$ be a measure space. Suppose $f \geq 0$ is measurable. Prove that $$\lim_{n \rightarrow \infty} \int{f \wedge n} \> d\mu = \int{f} d\mu.$$ (Here, the author defines $f \wedge n (x) = \min\{f(x),n\}$ for each $n \in \mathbb{N}$.)
$\textbf{Attempt at Solution:}$ Let $\epsilon > 0$ and let us suppose $ \int{f} d\mu < \infty$. We note the following:
$(1)$ $(f \wedge n)$ converges to $f$ pointwise (for each $x$ find an $N$ so large that $f(x) \leq N$),
$(2)$ for each $n \in \mathbb{N}$, $f \wedge n \leq f$ for all $x \in X$,
$(3)$ $(f \wedge n)$ is monotonically increasing and, therefore, $(\int{f \wedge n} \> d\mu)$ is monotonically increasing.
Item $(2)$ imples $\int{f \wedge n} \> d\mu \leq \int{f} d\mu$ for each $n$. The value $\int{f} d\mu - \epsilon$ is not an upper bound for the set $\{ \int{\phi} \> d\mu : \phi \text{ is simple, measurable and } 0 \leq \phi \leq f\}$. Hence, there exists some simple, measurable function $\phi_{0}$ such that $\int{f} d\mu - \epsilon < \int{\phi_{0}} \> d\mu \leq \int{f} d\mu$. Suppose $\{y_{1}, \ldots, y_{n}\}$ is the range for $\phi_{0}$. Put $N=\max\{y_{1}, \ldots, y_{n}\}$. Then, $\phi_{0} \leq f \wedge n$ for $n > N$. Consequently, $n > N$ implies $$\int{f} d\mu - \epsilon < \int{\phi_{0}} \> d\mu \leq \int{f \wedge n} \> d\mu \leq \int{f} d\mu < \int{f} d\mu + \epsilon.$$ That is, $\lim_{n \rightarrow \infty} \int{f \wedge n} \> d\mu = \int{f} d\mu.$
If $ \int{f} d\mu = \infty$, our desired results follows from Items (1) and (3).*
Is the above solution attempt correct? Thank you.
*I really want to be more explicit here. Let $M > 0$ be arbitrary. Assume $ \int{f} d\mu = \infty$. By definition, $\int{f} \> d\mu = \sup \{ \int{\phi} \> d\mu : \phi \text{ is simple, measurable and } 0 \leq \phi \leq f\}$. So, there must exist some simple, measurable function, say $\phi_{M}$, such that $M < \int{\phi_{M}} \> d\mu$. As in the previous case, suppose $\{y_{1}, \ldots, y_{n}\}$ is the range for $\phi_{M}$. Put $N=\max\{y_{1}, \ldots, y_{n}\}$. Then, $\phi_{M} \leq f \wedge n$ for $n > N$. Consequently, $M < \int{f \wedge n} \> d\mu $ for $n > N$. Because $M$ was arbitrary, $\int{f \wedge n} \> d\mu \rightarrow \infty$ as $n \rightarrow \infty$.
Looks fine. Just a small comment on your string of inequalities following "$n > N$ implies ...": I would only keep the first two inequalities, the latter two are not necessary.