# 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 $\int f d\lambda=I$

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I think everything is fine. First, you can prove wat you want for positive continuous functions, since $f$ continous implies $f^+$ and $f^-$ continuos and the integrables over $[0,1]$. Note that for fixed $n$, given $x\in [0,1]$, $x$ must belong to some $[\frac{i}{n},\frac{i+1}{n}]$ in that case by definition $h_n(x)\leq f(x)$. This proves $h_n\leq f$ for all $n$. The monotony of the sequence $(h_n)$ follows because the intervals considered to build $h_{n+1}$ are subintervals of some of the intervals used to build $h_n$, so it follows by properties of the infimum. –  leo Sep 22 '12 at 0:55
Finally $h_n\to h$. To prove this remember that since $f$ is continuous:$$\begin{array}{l}\text{1.it's uniformly continuous and }\\ \text{2.attains its extremums}\end{array}$$ over kompact intervals. –  leo Sep 22 '12 at 0:59

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.

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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.
Well the question is, do I now construct two sequences? To me it seems I'm asked to construct only one sequence and so far I am not able to find this sequence. For the positive parts of $f$ I agree it works to take $$h_{n}=\sum_{i=1}^{n}\min_{x \in [\frac{i}{n},\frac{i+1}{n}}f(x)\textbf{1}_{[\frac{i}{n},\frac{i+1}{n}]}$$ but then what for the negative part of $f$?? –  Math Girl Sep 22 '12 at 5:32
As Leo seems to hint at, your definition of $h_{n}$ works for the negative part too (if you remove the absolute value). –  Quinn Culver Sep 22 '12 at 12:48
Ok I understand that the sequence without absolute value is the right one. Now does anyone have any idea of how to prove the last part, $\int f d\lambda =I$?? We know that the sequence $h_{n}$ goes to f and thelebesgue integral of $h_{n}$ goes to I does $\int fd\lambda = I$ now follow from a theorem? –  Math Girl Sep 22 '12 at 17:05
I've considered this before but as this one only holds for simple functions with positive coefficient it does not hold here ($\min_{x\in[\frac{i}{n},\frac{i+1}{n}]}f(x)$ is not positive when we go into the negative part of $f$) –  Math Girl Sep 23 '12 at 16:22