Stream＃concat也很有用

Set<Long> SUBSCRIBER_ALLOWED_NUMBER_CD  =  
         Stream.concat(ORG_SUBSCRIBER_ALLOWED_NUMBER_CD.stream(),INDIVIDUAL_SUBSCRIBER_ALLOWED_NUMBER_CD.stream())
                .collect(Collectors.toSet());

如果要使用流：

public static final Set<Long> SUBSCRIBER_ALLOWED_NUMBER_CD =
    Stream.of(ORG_SUBSCRIBER_ALLOWED_NUMBER_CD,INDIVIDUAL_SUBSCRIBER_ALLOWED_NUMBER_CD)
        .flatMap(Set::stream)
        .collect(Collectors.toSet());

或者，就您的情况而言，如注释中所述，您可以使用Collectors.toUnmodifiableSet()：

public static final Set<Long> SUBSCRIBER_ALLOWED_NUMBER_CD =
    Stream.of(ORG_SUBSCRIBER_ALLOWED_NUMBER_CD,INDIVIDUAL_SUBSCRIBER_ALLOWED_NUMBER_CD)
        .flatMap(Set::stream)
        .collect(Collectors.toUnmodifiableSet());

只需指出，与仅JDK的解决方案相比，Guava处理此案的可读性更高（并且性能可能更高）：

public static final Set<Long> SUBSCRIBER_ALLOWED_NUMBER_CD = ImmutableSet.builder()
    .addAll(ORG_SUBSCRIBER_ALLOWED_NUMBER_CD)
    .addAll(INDIVIDUAL_SUBSCRIBER_ALLOWED_NUMBER_CD)
    .build();

您为什么不能创建第三组，而只添加两个最终组？

Set<Long> SUBSCRIBER_ALLOWED_NUMBER_CD = new HashSet<>();
SUBSCRIBER_ALLOWED_NUMBER_CD.addAll(ORG_SUBSCRIBER_ALLOWED_NUMBER_CD);
SUBSCRIBER_ALLOWED_NUMBER_CD.addAll(INDIVIDUAL_SUBSCRIBER_ALLOWED_NUMBER_CD);

前两个集合为final仅表示您不能修改那些集合。它不会阻止您将集合读入另一个新集合。

单语句初始化，如果需要的话：

public static final Set<Long> SUBSCRIBER_ALLOWED_NUMBER_CD = 
     Collections.unmodifiableSet(
        Stream.of(ORG_SUBSCRIBER_ALLOWED_NUMBER_CD,INDIVIDUAL_SUBSCRIBER_ALLOWED_NUMBER_CD)
              .flatMap(Set::stream)
              .collect(Collectors.toSet()));

或者，也许更具可读性：

public static final Set<Long> SUBSCRIBER_ALLOWED_NUMBER_CD;
static {
    Set<Long> all = new HashSet<>();
    all.addAll(ORG_SUBSCRIBER_ALLOWED_NUMBER_CD);
    all.addAll(INDIVIDUAL_SUBSCRIBER_ALLOWED_NUMBER_CD);
    
    SUBSCRIBER_ALLOWED_NUMBER_CD = Collections.unmodifiableSet(all);
}

如果预计第三组不可修改，则忽略Collections.unmodifiableSet调用。