Let $f:\left[0,1\right)\rightarrow\mathbb{R}$ monotonically increasing non-negative function s.t. $\int_{0}^{1}f\left(x\right)dx$ converge.
- Prove that for every $2\leqslant n\in\mathbb{N}$
$$\sum_{k=0}^{n-2}f\left(\frac{k}{n}\right)\cdot\frac{1}{n}\leqslant\int_{0}^{1}f\left(x\right)dx$$
- Let $0<\delta<1$. Prove that
$$\liminf_{n\rightarrow\infty}\sum_{k=0}^{n-2}f\left(\frac{k}{n}\right)\cdot\frac{1}{n}\geqslant\int_{0}^{1-\delta}f\left(x\right)dx$$
- Conclude that
$$\lim_{n\rightarrow\infty}\sum_{k=0}^{n-2}f\left(\frac{k}{n}\right)\cdot\frac{1}{n}=\int_{0}^{1}f\left(x\right)dx$$
Thoughts:
Using lower Darboux sum with partition $P=\left\{ 0,\frac{1}{n},\frac{2}{n},\dots\frac{n-1}{n}=1-\frac{1}{n}\right\} $, the answer is almost directly from definition, But I can't understand why the summation is from $0$ to $n-1$ and not to $n-2$, and what is the impact of this change on the sum.
Also, any direction for understanding the meaning of $\liminf$ as the the function is not defined?
Thank You.