# Prove $f:[a,b] \rightarrow \mathbb{R}$ is integrable and has zero content.

In my text book, it denotes the upper Riemann sum and lower Riemann sum as follow. $$S_Pf = \sum_{i=1}^n \sup\{f(x_i) : x_i \in [x_{i} - x_{i-1}]\}*(x_i - x_{i-1})$$ is the upper Riemann sum and $$s_Pf = \sum_{i=1}^n \inf\{f(x_i) : x_i \in [x_{i} - x_{i-1}]\}*(x_i - x_{i-1})$$ is the lower Riemann sum on some partition $P$.

The definition of zero content is a set $Z \subset \mathbb{R}^2$ has zero content if $\forall \epsilon > 0,$ there is a finite collection of rectangles $R_1, \ldots , R_n$ such that $Z \subset \bigcup_1^n R_i$ and the sum of the areas of the rectangles is less than $\epsilon.$

The question is for given integrable function $f:[a,b] -> \mathbb{R}$, show that the graph of $f$ in $\mathbb{R}^2$ has zero content or measure zero.

The following is what I have tried.

Let $P = \{a = x_0 < \cdots < x_n = b\}$ be a partition with fixed length $\frac{b-a}{n}$ and $M_I = \sup\{f(x_i) : x_i \in [x_{i} - x_{i-1}]\}$ and $m_i = \inf\{f(x_i) : x_i \in [x_{i} - x_{i-1}]\}.$ Then, we can define the upper and lower Riemann sum as $$S_Pf = \sum_{i=1}^n M_i \frac{b-a}{n}$$ $$s_Pf = \sum_{i=1}^n m_i \frac{b-a}{n}$$ Then, $S_Pf - s_Pf = \frac{b-a}{n} \left(\sum_{i=1}^n M_i - \sum_{i=1}^n m_i \right) < \epsilon, \forall \epsilon > 0.$ Is this acceptable?? The last step, there is a theorem in my text book that says if a function is integrable, then the difference between upper Riemann sum and lower Riemann sum is less than some positive $\epsilon.$ Thanks!

Let $\epsilon > 0$ be given. The integrability of $f$ allows to choose a partition $P$ of $[a,b]$ such that

$$S_Pf - s_Pf < \epsilon$$

Say $I$ is a subinterval determined by the partition $P$ and write

$$S_Pf - s_Pf = \sum_I ( \sup_{x \in I} f( x) - \inf_{x \in I} f(x) ) Lentgh(I)$$

It is easy to see that (Denote $\Gamma_f = \{ (x, f(x) ) \in \mathbb{R} \times \mathbb{R} : x \in [a,b] \}$ ) that

$$\Gamma_f \subset I \times [\inf_{x \in I} f, \sup_{x \in I} f] = \mathcal{C}$$

Hence, $\mathcal{C}$ provides a cover for the graph of $f$ and finally

$$\sum Area ( \mathcal{C} ) = \sum_I ( \sup_{x \in I} f( x) - \inf_{x \in I} f(x) ) Lentgh(I) < \epsilon$$