Regularity of measure in Lemma 7.2.6 of Bogachev

Question

In the book "Measure Theory" of Bogachev, vol. 2, Lemma 7.2.6 states the following.

Let $\mu$ be a $\tau$-additive, regular Borel measure on a topological space $X$, and let $\{f_\alpha\}$ be an increasing net of lower-semicontinuous non-negative functions such that $f:= \lim_\alpha f_\alpha$ is bounded. Then $$ \lim_\alpha \int_X f_\alpha \, d\mu \quad = \quad \int_X f \, d\mu . $$

In the proof it is quite clear where the $\tau$-additivity comes into play. However, I cannot see why we need $\mu$ to be regular. Bogachev defines this property in the following way (paraphrasing its Definition 7.1.5):

A measure $\mu$ on a topological space $X$ is called regular if for every measurable set $A\subseteq X$ and for every $\varepsilon > 0$ there exists a closed set $C_\varepsilon \subseteq A$ such that $\mu(A) - \mu(C_\varepsilon) < \varepsilon$.

Is regularity of $\mu$ in Lemma 7.2.6 then really necessary?

Answer 1

I found an authoritative reference: "Measure Theory" by D. H. Fremlin, vol. 4. Point a) of Corollary 414B therein says the following (the notes in the brackets are mine):

Let $X$ be a topological space and $\mu$ an effectively locally finite (for example, finite) $\tau$-additive topological (for example, Borel) measure on $X$. Suppose that $A$ is a non-empty upwards-directed family of lower-semicontinuous functions from $X$ to $[0.\infty]$. Set $$ g(x) := \sup_{f\in A} f(x) $$ for all $x\in X$. Then $$ \int g \, d\mu = \sup_{f\in A} \int f\,d\mu . $$

Modulo the difference in the notation, this implies the assert in the question.