Theorem 5.8.41A391017E17A

Proof. We first treat finite nonnegative functions, in Steps 1–3, and then reduce the general case to this one by truncation in Step 4. For the first part we prove by induction on $n\geq 1$ the following assertion: if $\Omega \subset \mathbb {C}^ n$ is a domain and $u\colon \Omega \to [0,\infty )$ is a finite function whose restriction to each connected component of each affine complex-line section of $\Omega $ is subharmonic, then $u$ is upper semicontinuous. For $n=1$ the only affine complex line is $\mathbb {C}$, so $u$ is subharmonic on $\Omega $ and in particular upper semicontinuous.

Let $n\geq 2$, assume the assertion in dimension $n-1$, and let $u$ be as in the assertion. Write $z=(z',w)$ with $z'\in \mathbb {C}^{n-1}$ and $w\in \mathbb {C}$. For fixed $w$, the restriction of $u(\, \cdot \, ,w)$ to each connected component of the open set $\{ z':(z',w)\in \Omega \} $ satisfies the hypothesis in dimension $n-1$. Indeed, affine complex lines in $\mathbb {C}^{n-1}\times \{ w\} $ are affine complex lines in $\mathbb {C}^ n$, a connected component of a line section of the slice lies in a connected component of the corresponding line section of $\Omega $, and subharmonicity is inherited by open subsets. By the induction hypothesis, $u(\, \cdot \, ,w)$ is therefore upper semicontinuous on this open set. For fixed $z'$, the function $u(z',\, \cdot \, )$ is subharmonic on each connected component of $\{ w:(z',w)\in \Omega \} $.

Step 1: $u$ is Borel measurable. We need joint measurability of $u$ before we can integrate it. Fix a ball $B'\subset \mathbb {C}^{n-1}$, a point $w_0\in \mathbb {C}$ and $\rho >0$ with $B'\times B(w_0,2\rho )\subset \Omega $. For integers $N\ge 1$ put $u_ N=\min \{ u,N\} $, and for $0<r<\rho $ and $(z',w)\in B'\times B(w_0,\rho )$ define

The integral exists, since $u(z',\, \cdot \, )$ is subharmonic, hence upper semicontinuous, on $B(w_0,2\rho )\supset B(w,r)$, so the integrand is a Borel function with values in $[0,N]$. For fixed $w$, the function $F_ r^ N(\, \cdot \, ,w)$ is upper semicontinuous: if $z'_ l\to z'$ in $B'$, the reverse Fatou lemma, applicable because $0\le u_ N\le N$, gives

where the last inequality holds because each $u_ N(\, \cdot \, ,\eta )$ is upper semicontinuous on $B'$. For fixed $z'$, the function $F_ r^ N(z',\, \cdot \, )$ is continuous: for $|w-v|\leq r$ the symmetric difference $B(w,r)\mathbin \triangle B(v,r)$ lies in the annulus $\{ r-|w-v|\leq |\eta -w|<r+|w-v|\} $, whose area is $4\pi r|w-v|$, so that $|F_ r^ N(z',w)-F_ r^ N(z',v)|\le 4N|w-v|/r$.

A function on $B'\times B(w_0,\rho )$ that is Borel in $z'$ and continuous in $w$ is Borel; we check this for $F_ r^ N$. For $k\geq 1$ partition $B(w_0,\rho )$ into countably many Borel sets $E_{k,i}$ of diameter less than $1/k$, choose a point $w_{k,i}$ in each nonempty $E_{k,i}$, and put $F_ k(z',w)=F_ r^ N(z',w_{k,i})$ for $w\in E_{k,i}$. Each $F_ k$ is Borel, because each $F_ r^ N(\, \cdot \, ,w_{k,i})$ is upper semicontinuous, hence Borel, and $F_ k(z',w)\to F_ r^ N(z',w)$ for every $(z',w)$, because $F_ r^ N(z',\, \cdot \, )$ is continuous and the chosen points $w_{k,i}$ with $w\in E_{k,i}$ tend to $w$. Thus $F_ r^ N$ is Borel, and so is $G_ N=\varlimsup _{k\to \infty }F^ N_{\rho /(k+1)}$.

Now fix $(z',w)\in B'\times B(w_0,\rho )$. The upper semicontinuous function $u(z',\, \cdot \, )$ is bounded above on the closed disc $\overline{B(w,\rho /2)}$ by some $M$. For $N\geq M$ the truncation does not change $u(z',\, \cdot \, )$ on this disc, so $F^ N_{\rho /(k+1)}(z',w)$ is the mean of $u(z',\, \cdot \, )$ over $B(w,\rho /(k+1))$. The function $u(z',\, \cdot \, )$ is subharmonic and finite on $B(w_0,2\rho )$, hence locally integrable there by proposition 3.4.8, and by (3.4.13) its means over the discs $B(w,\rho /(k+1))$ converge to $u(z',w)$ as $k\to \infty $. Hence $G_ N(z',w)=u(z',w)$ for all $N\geq M$, and $u=\lim _ NG_ N$ is Borel on $B'\times B(w_0,\rho )$. Countably many such product sets cover $\Omega $, so $u$ is Borel on $\Omega $.

Step 2: the canonical representative. By corollary 2.9.28, the finite Borel function $u$ is locally bounded above. Since $u\geq 0$, it follows that $u\in L^1_{\mathrm{loc}}(\Omega )$. Since the restriction of $u$ to each connected component of each affine complex-line section of $\Omega $ is subharmonic, lemma 5.8.2 gives $\mathrm{dd}^{\mathrm{c}}u\geq 0$. In particular $\Delta u=4\sum _ ju_{j\bar j}\geq 0$, so by theorem 3.4.12 $u$ has a canonical subharmonic representative $\widetilde u$, a locally integrable subharmonic function on $\Omega $ equal to $u$ almost everywhere. Its complex Hessian is the same positive current $\mathrm{dd}^{\mathrm{c}}u$. Applying proposition 5.4.4 to the parametrizations of affine complex lines, we see that the restriction of $\widetilde u$ to each connected component of each affine complex-line section of $\Omega $ is subharmonic or identically $-\infty $. Since $\widetilde u$ is upper semicontinuous and agrees almost everywhere with the finite function $u$, we conclude that $\widetilde u\in \mathrm{PSH}(\Omega )$.

Step 3: $u=\widetilde u$ everywhere. Fix $x\in \Omega $ and $r>0$ with $\overline{B(x,r)}\subset \Omega $. The set $Z=\{ u\ne \widetilde u\} $ is Borel, by Step 1, and Lebesgue null, so the left-hand side of (5.8.3) vanishes for $F=\mathbf1_ Z$. Since the weight $|t|^{2n-2}$ is positive for $t\neq 0$, for $\tau $-almost every $a$ the set $\{ t:|t|<r,\ x+ta\in Z\} $ has planar measure zero. Fix such an $a$. On the disc $\{ |t|<r\} $, the function $t\mapsto u(x+ta)$ is finite and subharmonic, while $t\mapsto \widetilde u(x+ta)$ is subharmonic or identically $-\infty $; the latter is impossible, because the two functions agree almost everywhere on the disc. By proposition 3.4.8 both are locally integrable subharmonic functions on the disc that agree almost everywhere, so they agree everywhere by theorem 3.4.12. At $t=0$ this gives $u(x)=\widetilde u(x)$. As $x$ was arbitrary, $u=\widetilde u$ is upper semicontinuous. This completes the induction.

Step 4: truncation. Let $u$ be as in the theorem, and for integers $m\geq 1$ put $u_ m=\max \{ u,-m\} +m$. This is a finite nonnegative function. Its restriction to each connected component of each affine complex-line section of $\Omega $ is subharmonic: if the restriction of $u$ is subharmonic, so is its maximum with the constant $-m$, and if the restriction of $u$ is identically $-\infty $, the restriction of $u_ m$ is the constant $0$. By the case already proved, each $u_ m$, and hence each $\max \{ u,-m\} $, is upper semicontinuous. For every real $c$ we have $\{ u<c\} =\bigcup _ m\{ \max \{ u,-m\} <c\} $, which is open. Thus $u$ is upper semicontinuous, and together with the hypotheses this gives $u\in \mathrm{PSH}(\Omega )$. $\square$