following two lemmas are stated in Terence Tao's book "an introduction to measure theory" and proof is left as an exesice:
(Horizontal truncation) As ${n \rightarrow \infty}, {\underline{\int_{{\bf R}^d}} \min(f(x),n)\ dx}$ converges to ${\underline{\int_{{\bf R}^d}} f(x)\ dx}$.
(Vertical truncation) As ${n \rightarrow \infty}, {\underline{\int_{{\bf R}^d}} f(x) 1_{|x| \leq n}\ dx}$ converges to ${\underline{\int_{{\bf R}^d}} f(x)\ dx}.$
I have proved both, but I'm not quite sure they are valid proofs.
(Horizontal truncation proof): $$\lim_{n\to \infty} {\underline{\int_{{\bf R}^d}} \min(f(x),n)\ dx} = \lim_{n\to \infty} \sup_{\psi(x) \leq min(f(x),n)} \underline{\int_{{\bf R}^d}}\psi(x)dx\ \ \ \psi \ simple = $$ $$\sup_{\psi(x) \leq f(x)} \underline{\int_{{\bf R}^d}}\psi(x)dx\ \ \ \psi \ simple = \underline{\int_{{\bf R}^d}}f(x)dx$$ (Vertical truncation proof): $$\lim_{n\to\infty}{\underline{\int_{{\bf R}^d}} f(x) 1_{|x| \leq n}\ dx} = \lim_{n\to \infty} \sup_{\psi(x) \leq f(x)1_{|x| \leq n}} \underline{\int_{{\bf R}^d}}\psi(x)dx\ \ \ \psi \ simple =$$ $$\sup_{\psi(x) \leq f(x)} \underline{\int_{{\bf R}^d}}\psi(x)dx = \underline{\int_{{\bf R}^d}}f(x)dx$$
The main problem is that I'm not sure taking n inside \sup is valid or not.
If you find proofs wrong I'll be really thankful if you leave a correct proof and also counterexamples for where my argument goes wrong.Thanks.
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.