Lemma 3.4.436C9571A59E0

Proof. Since $f$ is bounded above, we can choose a positive integer $n_0>\sup _ Xf$. For $n\geq n_0$ and $x\in X$ put

Taking $y=x$ gives $f_ n(x)\geq \max \{ f(x),-n\} \geq f(x)$, and $f_ n\leq \max \{ \sup _ Xf,-n\} <n_0$. Thus $-n\leq f_ n<n_0$, so $f_ n$ is real-valued and bounded, and as a supremum of $n$-Lipschitz functions of $x$ it is $n$-Lipschitz. Both $\max \{ f(y),-n\} $ and $-n\operatorname {dist}(x,y)$ decrease as $n$ increases, so the sequence $(f_ n)$ is decreasing.

We show that $f_ n(x)\to f(x)$ for every $x\in X$. Otherwise there is a real number $c$ with $\lim _ nf_ n(x)>c>f(x)$. For each $n$ choose $y_ n\in X$ with $\max \{ f(y_ n),-n\} -n\operatorname {dist}(x,y_ n)>c$. Then $n\operatorname {dist}(x,y_ n)<n_0-c$, so $y_ n\to x$. Moreover $\max \{ f(y_ n),-n\} >c$, which forces $f(y_ n)>c$ as soon as $n>-c$. On the other hand, $\{ f<c\} $ is an open neighborhood of $x$ by upper semicontinuity, so it contains $y_ n$ for all large $n$. This contradiction proves that $f_ n\downarrow f$.

Finally, the nonnegative functions $n_0-f_ n$ increase to $n_0-f$, so the monotone convergence theorem gives $\int (n_0-f_ n)\, \mathrm d\mu \uparrow \int (n_0-f)\, \mathrm d\mu $. Subtracting these integrals from the finite number $n_0\mu (X)$ proves the assertion about the integrals of $f_ n$. $\square$