Let $\lambda$ denote the Lebesgue measure on the Borel sets of [0,1]. Let $f: [0,1] \rightarrow \mathbb{R}$ be continuous. I know that the Riemann integral $I:=\int_{0}^{1} f(x)dx$ exists. I also know that the Lebesgue integral of $f$ exists. The question is to construct an increasing sequence of simple functions $h_{n}$ with limit $f$ satisfying $h_{n}\leq f$ and $\int h_{n} \ d\lambda \ \uparrow I$.
The hint was to use the definition of the Riemann integral so I tried.. We know because $h_{n}$ needs to be simple that it is of the form $\sum_{i=1}^{n}\alpha_{i} \textbf{1}_{A_{i}}$. My idea for $h_{n}$ now was
$h_{n}=\sum_{i=1}^{n}\min_{x\in[\frac{i}{n},\frac{i+1}{n}]}|f(x)| \textbf{1} _{\{[\frac{i}{n},\frac{i+1}{n}]\}}$
If $s\in[\frac{i}{n},\frac{i+1}{n}]$ then $h_{n}(s)$ takes the minimum value of the function $|f|$ on this interval.
It is obvious that we get $\int h_{n} \ d\lambda=\sum_{i=1}^{n} \min_{x\in[\frac{i}{n},\frac{i+1}{n}]} |f(x)| \cdot \frac{1}{n}$
This indeed converges to $I$, the area under the function $f$ but now $h_{n}\leq f$ does not hold and neither is the sequence $h_{n}$ increasing to $f$.
I've also tried to not take the absolute value of $f$ in the function of $h_{n}$ but just the value $f(x)$ but the the lebesgue integral of $h_{n}$ does not go to the area under the function $f$.
Could anyone help me find such a sequence $h_{n}$??
I then also have to prove that the Lebesgue integral of $f$ is equal to the Riemann integral, so $\int f d\lambda=I$