I have a task in my textbook, which is the following:
For $f: [1,2] \rightarrow \Bbb R$ , $f(x)= 1/x$ choose a sequence of step functions $\phi_n$ approximating $f$ with partition $P_n= \{r/n:n \leq r \leq 2n\}$ to show that $$\frac{1}{n+1}+...+\frac{1}{2n} \rightarrow \int^2_1x^{-1}dx=log(2) \text { as $n\rightarrow \infty$}$$
What I don't really understand is how can I find a sequence of step functions with such a partition.
Since how I view this partition is :
For $n=1$ $P_1=\{r: 1\leq r \leq 2\}$
For $n=2$ $P_2=\{r/2 : 2\leq r \leq 4\}$
For $n=3$ $P_3=\{r/3: 3\leq r \leq 6\}$
And so on...
So how am I supposed to construct a sequence of step functions which approximates $f$ ? Since the values that $\phi_n$ can take are bigger than $1$. So for $P_1$ it will take minimum value $1$ (and so on for all the other $P_n$'s). I think I don't quite understand how $\phi_n$ is constructed from $P_n$ in the first place. I basically can't imagine how this step function could be something like $\frac{1}{1},\frac{1}{11},\frac{1}{111},...,\frac{1}{199999....},\frac{1}{2}$ which is how it's supposed to look like, right?
Just compute $L(f,P_n) = \sum_{r=n}^{2n-1} (\inf_{x \in [{ r \over n}, {r+1 \over n}]} f(x) ) ({r+1 \over n} - {r \over n})$.
Note that since $f$ is decreasing, $ \inf_{x \in [{ r \over n}, {r+1 \over n}]} f(x) = f({r+1 \over n}) = {1 \over {r+1 \over n}}$.
The step function required is $s_n = \sum_{r=n}^{2n-1} {n \over r+1} \cdot 1_{ [{ r \over n}, {r+1 \over n})} $