PomeProposition 16.28.3.181D96E0C805 If $f$ is bounded and lower semicontinuous, then $P_\omega (f)$ is bounded, lies below $f$ quasi-everywhere, and its Monge–Ampère measure is concentrated on $\{ P_\omega (f)=f\} $.
Proposition 16.28.3181D96E0C805
Proof. Suppose first that $f$ is continuous. Choose smooth functions $f_ j\to f$ uniformly. The Lipschitz bound for envelopes in proposition 5.40.3 gives $p_ j=P_\omega (f_ j)\to p=P_\omega (f)$ uniformly. Every compact subset of the open set $\{ p<f\} $ has a neighborhood on which $p_ j<f_ j$ for all large $j$. The preceding proposition and convergence of bounded products then give $\omega _ p^ n=0$ on $\{ p<f\} $.
For lower semicontinuous $f$, choose continuous functions $f_ j\uparrow f$ with common upper and lower bounds. For example, for a fixed distance $d$ inducing the topology of $X$, one may take
\[ f_ j(x)=\inf _{y\in X}\bigl (f(y)+j d(x,y)\bigr ). \]
These functions are Lipschitz and increase to $f$: if convergence failed at $x$, approximate minimizers $y$ would approach $x$, contradicting lower semicontinuity of $f$ at $x$.
The functions $p_ j=P_\omega (f_ j)$ increase and stay bounded, so the regularized limit $p_\infty =(\lim _ jp_ j)^*$ is a bounded $\omega $-plurisubharmonic function. It agrees with $\lim _ jp_ j$ outside a pluripolar set, and since each $p_ j$ lies below $f_ j\leq f$ outside a pluripolar set, we obtain $p_\infty \leq f$ quasi-everywhere. Thus $p_\infty \leq P_\omega (f)$.
For fixed $k$, the set $\{ p_\infty <f_ k\} $ is open, since $p_\infty $ is upper semicontinuous and $f_ k$ is continuous. If $j\geq k$, then $p_ j\leq p_\infty <f_ k\leq f_ j$ there, so the continuous case gives $\omega _{p_ j}^ n=0$ on that open set. Bedford–Taylor continuity along increasing sequences applies after adding a fixed smooth local function whose Hessian dominates $\omega $ and expanding with respect to the resulting smooth form. Together with the lower semicontinuity of the mass of open sets under weak convergence, it yields
\[ \omega _{p_\infty }^ n(\{ p_\infty <f_ k\} )=0. \]
Taking the union over $k$ gives zero measure on $\{ p_\infty <f\} $, hence also on $\{ p_\infty <P_\omega (f)\} $ outside a pluripolar set, since $P_\omega (f)\leq f$ quasi-everywhere. Bounded measures of this kind do not charge pluripolar sets, so the domination principle, theorem 14.19.10, gives $p_\infty \geq P_\omega (f)$. The two functions are therefore equal, and the assertion follows. $\square$
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