Let $f: [0,1] \to \mathbb{R}$ a strictly positive continuous function. Let $A = \{(x,y) \in \mathbb{R}^{2} \mid x \in [0,1], 0 \leq y \leq f(x)\}$. Whereas the Lebesgue Measure and the Riemann Integral, prove that $$m(A) = \int_{0}^{1}f(x)\mathrm{d}x.$$
My attemp. Let $P = \{t_{0} = 0 < t_{1} < \cdots < t_{n-1} < t_{n} = 1\}$ be a partition of $[0,1]$ and $\mathcal{P}$ the set of all partition of $[0,1]$. By definition $$\int_{0}^{1}f(x)\mathrm{d} x = \inf_{P \in \mathcal{P}}\sum M_{i}(t_{i} - t_{i-1}),$$ where $M_{i} = \sup_{[t_{i-1},t_{i}]}f(x)$. Also, by definition, $$m_{\ast}(A) = \inf_{Q \in \mathcal{C}} \sum |Q_{j}|,$$ where $Q = \bigcup Q_{j}$ is a cover of $A$ by closed cubes and $\mathcal{C}$ is the set of all convers of $A$ by closed cubes (or rectangles). Is easy to see that each $\sum M_{i}(t_{i} - t_{i-1})$ is a cover of $A$ by closed rectangles and with this, $$m_{\ast}(A) \leq \sum M_{i}(t_{i} - t_{i-1})$$ for any partition $P$ of $[0,1]$ and so, $$m_{\ast}(A) \leq \int_{0}^{1}f(x)\mathrm{d}x.$$ Now, we can analogously to use the definition of Riemann integral with the $\inf$, which clearly implies $$\int_{0}^{1}f(x)\mathrm{d}x \leq m_{\ast}(A)$$ since the sums of integral are inferior to the function $f$. Since $f$ integrable, we have $$m_{\ast}(A) \leq \int_{0}^{1}f(x)\mathrm{d}x \leq m_{\ast}(A)$$ and the result follows.
But, I'm not very sure with that idea. Can anybody help me?
Your idea is essentially correct. By using the Darboux approach to Riemann integral you can get your needed result. An upper Darboux sum $S_U(p)$ is the sum over a partition $(p)$ of the interval of rectangles where the height is given by the maximum of the function within each partition interval and a lower Darboux sum $S_L(p)$is the sum over a partition of the interval of rectangles where the height is given by the minimum of the function within each partition interval. For each $(p)$ we have $S_L(p)\le m(A),\ \int_0^1f(x)dx\le S_U(p)$. Since the function is Riemann integrable, the two Darboux bounds converge together to the Riemann integral as the partition maximum interval $\to 0$, so that the integral and the measure are squeezed together..