Python模块使用__main__.py运行时，子进程日志无法继承父进程配置的问题求助

我遇到了一个关于Python多进程日志的奇怪问题，先给大家看看我的最小复现环境：

目录结构 #

/mymodule
├── __init__.py
├── main.py
└── worker.py

其中__init__.py是空文件。

main.py代码 #

import sys
import logging
import multiprocessing
from test.worker import do_stuff

logging.basicConfig(
    format='[%(name)s] [%(levelname)s]: %(message)s',
    level=logging.DEBUG,
)

logger = logging.getLogger(__name__)

def main():

    logger.debug('I am main. I manage workers')
    logger.info('I am main. I manage workers')
    logger.warning('I am main. I manage workers')

    p = multiprocessing.Process(target=do_stuff)
    p.start()

if __name__ == '__main__':
    sys.exit( main() )

worker.py代码 #

import logging

logger = logging.getLogger(__name__)

def do_stuff():
    logger.debug(f'I am a worker. I do stuff')
    logger.info(f'I am a worker. I do stuff')
    logger.error(f'I am a worker. I do stuff')
    logger.error(f'Here is my logger: {logger}')

正常运行的情况 #

当我执行python -m mymodule.main时，输出完全符合预期：

[__main__] [DEBUG]: I am main. I manage workers
[__main__] [INFO]: I am main. I manage workers
[__main__] [WARNING]: I am main. I manage workers
[mymodule.worker] [DEBUG]: I am a worker. I do stuff
[mymodule.worker] [INFO]: I am a worker. I do stuff
[mymodule.worker] [ERROR]: I am a worker. I do stuff
[mymodule.worker] [ERROR]: Here is my logger: <Logger mymodule.worker (DEBUG)>

出现问题的情况 #

但我只是把/mymodule/main.py重命名为mymodule/__main__.py，然后执行python -m mymodule.__main__或者python -m mymodule时，输出就不对了：

[__main__] [DEBUG]: I am main. I manage workers
[__main__] [INFO]: I am main. I manage workers
[__main__] [WARNING]: I am main. I manage workers
I am a worker. I do stuff
Here is my logger: <Logger mymodule.worker (WARNING)>

很明显，第二种情况下mymodule.worker的logger没有继承logging.basicConfig的配置，不仅日志格式丢失了，日志级别还变成了WARNING。我完全没改代码，只是改了文件名而已，这是为什么呢？

我希望能通过__main__.py实现直接用模块名运行的便捷性，同时还能让logging.basicConfig的配置正确被子模块和子进程继承，该怎么解决这个问题呢？

备注：内容来源于stack exchange，提问作者Art Gertner