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On Example 1 in this following PDF: http://home.iitk.ac.in/~psraj/mth101/lecture_notes/lecture15-16.pdf

Consider the function $f : [0, 1] \to \mathbb R$ defined by $f(1/2) = 1$ and $f(x) = 0$ for all $x \in [0, 1] \setminus \{1/2\}$

I am trying to accumulate alot of different answered questions in real analysis so that I can start to become more familiar with proofs. I found this PDF today which seems like a pretty good overview (since we are covering integration in class) but on the sheet he says that "For any partition $P$ of $[a,b]$, the $L(f,P)$ is $0$". How do we know this?

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    No worries. I copied the example to the question this time. :)2012-10-29

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The function $f$ is $0$ everywhere in $[0, 1]$ except for $1/2$. Thus, no matter what partition you choose, the infimum of $f$ will be $0$ in each interval in this partition. This is clear when for the intervals that don't contain $1/2$. For the interval that does contain $1/2$, it must also contain another point, and hence the infimum is also $0$.

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    Note you could also define a partition of $[a,b]$ to be an ordered tuple $(x_0,x_1,x_2,\dotsc,x_n)$ with $a=x_0 \leq x_1 \leq \dotsb \leq x_n = b$, i.e. some of the $x_i$ could be equal. Then you could have $\tfrac12$ show up multiple times in your tuple at, say, $x_i$ and $x_{i+1}$ and thus $[x_i,x_{i+1}]$ wouldn't contain another point, but this isn't a problem as the length of the interval $[x_i,x_{i+1}]$ is $0$, so it doesn't contribute anything to $L(f,P)$2012-10-29