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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)?

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Alternative approach.

Suppose that the statement is false.

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.

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  • $\begingroup$ This works as well. And if $g = \infty$ on a set of positive measure, then clearly $\lim a_n = \int f = \infty$. $\endgroup$
    – shark
    Commented Feb 22, 2023 at 17:18

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