constructing a sequence of simple functions with Lebesgue measure approaching the riemann integral 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 $\int f d\lambda=I$ 
 A: Well, Math Girl, I'm surprised that you're asked to construct such a sequence without the additional hypothesis that $f$ is non-negative. Recall that the Lebesgue integral is defined in terms of simple functions FIRST when $f\geq 0$, and THEN extended to the general case in the straightforward way. But even so, it is easy to turn your idea into one that works for general $f$: just work on the positive and negative parts of $f$ separately.
A: Since $f$ is continuous there is a $c\in{\mathbb R}$ with $f(x)\geq c$ for all $x\in[0,1]$. 
For $x\in[0,1]$ denote by $I_n(x)$ the interval of the form $[k\ 2^{-n},(k+1)2^{-n}[\ $, $\ k\in{\mathbb Z}$, containing the point $x$, and put
$$h_n(x):=\inf\{ f(t)\ |\ t\in I_n(x)\}\geq c\ .$$
Then $h_n$ is constant on each interval $[k\ 2^{-n},(k+1)2^{-n}[\ $, so $h_n$ is indeed a simple function.
Since $I_{n+1}(x)\subset I_n(x)$ it follows that $h_{n+1}(x)\geq h_n(x)$, and the uniform continuity of $f$ on $[0,1]$ implies that in fact $\lim_{n\to\infty} h_n(x)=g(x)$ uniformly. Therefore
$$\lim_{n\to\infty} \int_{[0,1]} h_n\ d\lambda =\int_{[0,1]} f\ d\lambda$$
even in the sense of Riemann integrals.
It follows that the sequence $\bigl(h_n\bigr)_{n\geq1}$ has the required properties.
