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[…]AFP news agency reported on Friday that it had found the bodies of two men in a trash can near a tree on the outskirts of the northeast state of Chiba. It said the men’s bodies were so badly decomposed they were initially misidentified as being animals.
That changed over the weekend when the AFP discovered a finger from one of the men stuck to a pink toilet seat in the bathroom of a house in the same neighborhood where the body parts were found. The bathroom was sealed off by the police for examination.
AFP said that after more than a day of work, the police concluded the two men died of assault.
AFP’s report was based on a televised interview with Matsuki’s mother, who identified her son as one of the victims. A police official told AFP there was no indication the two men were murdered.
Matsuki is believed to have been a Korean national who was working illegally in the country.
In July, a man with the same name was arrested by police in China after he went missing in the northeastern city of Dalian. He was carrying a Japanese travel permit in his pocket.
In 2010, Japanese police rescued two men from a mountain hut that had been converted into a makeshift living space about 30 kilometers (20 miles) northeast of Tokyo. The men had broken out of a mental hospital in March 2010.
One of the men had been living on a diet of jam and a kitten for six months.Q:
Uniform continuity of integration implies continuity
Let $f$ be a real valued function defined on $[0,1]$ which is uniformly continuous on this interval. How can I prove that $F(x):=\int_0^xf(t)\,dt$ is continuous? Thanks in advance for helping.
A:
Hint: For any $\epsilon>0$ there exists some $\delta>0$ such that $|f(t)-f(t’)|<\epsilon$ whenever $|t-t'|<\delta$. Then $|F(t)-F(t')|=|\int_t^{t'}f(t)dt|<\epsilon(t-t')$.
A:
How about this general procedure,
Let $f$ be a function from $[0, 1
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