You will need the dominated convergence theorem in order to justify the interchange of the limit and the integration operator. Instead, I will state the more general form of the horizontal\vertical truncation properties which appear later in Exercise $1.4.36$ of the same textbook.

Let $(X, \mathcal{B}, \mu)$ be a measure space, and let $f$, $g$: $X \rightarrow [0, +\infty]$ be measurable.

(Horizontal truncation) We have $\lim_{n \to \infty} \int_X \min(f, n)\ d\mu = \int_X f\ d\mu$.

(Vertical truncation) If $E_1 \subset E_2 \subset \ldots$ is an increasing sequence of $\mathcal{B}-$measurable sets, then

$\lim_{n \to \infty}\int_X f1_{E_n}\ d\mu = \int_X f1_{\cup_{n=1}^\infty E_n}$.

Proof of the horizontal truncation:

First assume that $\int_X f\ d\mu < \infty$, then $f$ is finite almost everywhere. In particular, the functions $f_n = \min(f, n)$ agree with $f$ almost everywhere for sufficiently large $n$. Thus $\lim_{n \to \infty} \int_X \min(f, n)\ d\mu = \int_X f\ d\mu$.

Next assume that $\int_X f\ d\mu = \infty$. Then either $f = +\infty$ on a set $E$ of positive measure, or that the set $\{f > 0\}$ has infinite measure.

In the former case, pick $M > 0$. If we let $g$ be the simple function define by $g := M1_E$, then $\int_X \min(f, M)\ d\mu \geq$ Simp $\int_X g\ d\mu \geq M\mu(E)$. Sending $M \rightarrow \infty$, and using monotonicity of the unsigned integral, we get $\lim_{M \to \infty} \int_X \min(f, M)\ d\mu = \infty$.

In the latter case, we can further assume that $f < \infty$ almost everywhere. Then again for sufficiently large $M$, we have by equivalency that $\int_X \min(f, M)\ d\mu = \int_X f\ d\mu = \infty$. That is, $\lim_{M \to \infty} \int_X \min(f, M)\ d\mu = \infty$, as desired.

Proof of the vertical truncation:

First assume that $f$ is simple. Then $f = \sum_{i=1}^n a_i1_{A_i}$, a finite combination of indicators of measurable sets. Let $E = \bigcup_{n=1}^\infty E_n$. For any natural number $m$, we have

$|\text{Simp} \int_X f1_E\ d\mu\ -\ \text{Simp} \int_X f1_{E_m}\ d\mu| = |\sum_{i=1}^n a_i \mu(E \cap A_i) - \sum_{i=1}^n a_i \mu(E_m \cap A_i)|$

$= \sum_{i=1}^n a_i(\mu(E \cap A_i) - \mu(E_m \cap A_i))$.

Note that for each $i$, $(E_1 \cap A_i) \subset (E_2 \cap A_i) \subset \ldots$ is an increasing sequence of measurable sets. Taking $m \rightarrow \infty$ in the above identity and using continuity from below (upwards monotone convergence) give the claim for the case when $f$ is simple.

Now for a general unsigned measurable $f$. By definition, we can find a simple function $g \leq f$ such that $|\int_X f\ d\mu\ -\ \text{Simp} \int_X g\ d\mu| \leq \varepsilon$, for any $\varepsilon > 0$. By monotonicity and the triangle inequality, we then have

$|\int_X f\ d\mu\ - \int_X f1_{f \geq 1/n}| \leq |\int_X f\ d\mu - \int_X f1_{g \geq 1/n}\ d\mu| \leq |\int_X f\ d\mu - \int_X g1_{g \geq 1/n}\ d\mu|$

$\leq |\int_X f\ d\mu - \int_X g\ d\mu| + |\int_X g\ d\mu - \int_X g1_{g \geq 1/n}\ d\mu| \leq \varepsilon + |\int_X g\ d\mu - \int_X g1_{g \geq 1/n}\ d\mu|$.

Taking $n \rightarrow \infty$, and then $\varepsilon \rightarrow 0$ gives the desired result.

You will need the dominated convergence theorem in order to justify the interchange of the limit and the integration operator. Instead, I will state the more general form of the horizontal\vertical truncation properties which appear later in Exercise $1.4.36$ of the same textbook, and prove them without using dominated convergence.

Let $(X, \mathcal{B}, \mu)$ be a measure space, and let $f$, $g$: $X \rightarrow [0, +\infty]$ be measurable.

(Horizontal truncation) We have $\lim_{n \to \infty} \int_X \min(f, n)\ d\mu = \int_X f\ d\mu$.

(Vertical truncation) If $E_1 \subset E_2 \subset \ldots$ is an increasing sequence of $\mathcal{B}-$measurable sets, then

$\lim_{n \to \infty}\int_X f1_{E_n}\ d\mu = \int_X f1_{\cup_{n=1}^\infty E_n}$.

Proof of the horizontal truncation:

First assume that $\int_X f\ d\mu < \infty$, then $f$ is finite almost everywhere. In particular, the functions $f_n = \min(f, n)$ agree with $f$ almost everywhere for sufficiently large $n$. Thus $\lim_{n \to \infty} \int_X \min(f, n)\ d\mu = \int_X f\ d\mu$.

Next assume that $\int_X f\ d\mu = \infty$. Then either $f = +\infty$ on a set $E$ of positive measure, or that the set $\{f > 0\}$ has infinite measure.

In the former case, pick $M > 0$. If we let $g$ be the simple function define by $g := M1_E$, then $\int_X \min(f, M)\ d\mu \geq$ Simp $\int_X g\ d\mu \geq M\mu(E)$. Sending $M \rightarrow \infty$, and using monotonicity of the unsigned integral, we get $\lim_{M \to \infty} \int_X \min(f, M)\ d\mu = \infty$.

In the latter case, we can further assume that $f < \infty$ almost everywhere. Then again for sufficiently large $M$, we have by equivalency that $\int_X \min(f, M)\ d\mu = \int_X f\ d\mu = \infty$. That is, $\lim_{M \to \infty} \int_X \min(f, M)\ d\mu = \infty$, as desired.

Proof of the vertical truncation:

First assume that $f$ is simple. Then $f = \sum_{i=1}^n a_i1_{A_i}$, a finite combination of indicators of measurable sets. Let $E = \bigcup_{n=1}^\infty E_n$. For any natural number $m$, we have

$|\text{Simp} \int_X f1_E\ d\mu\ -\ \text{Simp} \int_X f1_{E_m}\ d\mu| = |\sum_{i=1}^n a_i \mu(E \cap A_i) - \sum_{i=1}^n a_i \mu(E_m \cap A_i)|$

$= \sum_{i=1}^n a_i(\mu(E \cap A_i) - \mu(E_m \cap A_i))$.

Note that for each $i$, $(E_1 \cap A_i) \subset (E_2 \cap A_i) \subset \ldots$ is an increasing sequence of measurable sets. Taking $m \rightarrow \infty$ in the above identity and using continuity from below (upwards monotone convergence) give the claim for the case when $f$ is simple.

Now for a general unsigned measurable $f$. By definition, we can find a simple function $g \leq f$ such that $|\int_X f\ d\mu\ -\ \text{Simp} \int_X g\ d\mu| \leq \varepsilon$, for any $\varepsilon > 0$. By monotonicity and the triangle inequality, we then have

$|\int_X f\ d\mu\ - \int_X f1_{f \geq 1/n}| \leq |\int_X f\ d\mu - \int_X f1_{g \geq 1/n}\ d\mu| \leq |\int_X f\ d\mu - \int_X g1_{g \geq 1/n}\ d\mu|$

$\leq |\int_X f\ d\mu - \int_X g\ d\mu| + |\int_X g\ d\mu - \int_X g1_{g \geq 1/n}\ d\mu| \leq \varepsilon + |\int_X g\ d\mu - \int_X g1_{g \geq 1/n}\ d\mu|$.

Taking $n \rightarrow \infty$, and then $\varepsilon \rightarrow 0$ gives the desired result.