# vertical truncation property for the unsigned integral

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

(Vertical truncation) Show that $${\lim_{n \rightarrow \infty} \int_X \min(f,n)\ d\mu = \int_X f\ d\mu}$$.

The hint is that first try the case when $$f$$ is a simple function. I was able to establish that. I think the idea then is that for a general measurable $${f: X \rightarrow [0,+\infty]}$$, $$\int_X f\ d\mu := \sup_{0 \leq g \leq f, g \text{ simple}}\int_X g\ d\mu = \sup_{0 \leq g \leq f, g \text{ simple}}(\lim_{n \rightarrow \infty} \int_X \min(g,n)\ d\mu) \stackrel{*}{=} \lim_{n \rightarrow \infty}(\sup_{0 \leq g \leq f, g \text{ simple}}\int_X \min(g,n)\ d\mu) = \lim_{n \rightarrow \infty}(\sup_{0 \leq h \leq \min(f,n), h \text{ simple}}\int_X h d\mu) := \lim_{n \rightarrow \infty}\int_X \min(f,n) d\mu$$.

Question: On what ground can we justify $$(*)$$(the swapping of the limit with the supremum)?

If $$a_n:=\int\min(f,n)$$ then $$(a_n)_n$$ is an increasing sequence in $$[0,\infty]$$ so if it does not converge to $$\int f$$ then some $$c\in[0,\infty)$$ must exist with:$$a_n\leq c<\int f\text{ for every }n$$
Then a simple function $$g$$ exists with $$g\leq f$$ and $$c<\int g$$.
But because $$g$$ is simple for $$n$$ large enough we have $$g=\min(g,n)$$ so that: $$c<\int g=\int\min(g,n)\leq a_n\leq c$$ A contradiction.
• This works as well. And if $g = \infty$ on a set of positive measure, then clearly $\lim a_n = \int f = \infty$. Commented Feb 22, 2023 at 17:18