Proof: Define
, and let
(resp.
) be arbitrary. Let
and
. Since
is locally finite, pick a neighbourhood
of
such that
. Since
is continuous, by shrinking
if necessary, we may assume that for
we have
. Since
is compact, we may choose
so that
. Now for each arbitrary finite open cover
of
and
for
define the distribution
,
這實際上是一個所需型別的分佈 (
或
。在上面構建的覆蓋的特定情況下,請注意
.
進一步注意,型別為
的元組,其中
且
是
的一個開覆蓋,在以下關係下構成一個有向集
,
根據上述計算,
的淨值逐點收斂於
。由於來自桶形 LCTVS 到 Hausdorff TVS 的連續線性函式的逐點極限是連續且線性的,我們得出結論。