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 $\int f d\lambda=I$

  • $\begingroup$ 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. $\endgroup$ – leo Sep 22 '12 at 0:55
  • $\begingroup$ 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. $\endgroup$ – 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.


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.

  • $\begingroup$ 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 $f$?? $\endgroup$ – Math Girl Sep 22 '12 at 5:32
  • $\begingroup$ As Leo seems to hint at, your definition of $h_{n}$ works for the negative part too (if you remove the absolute value). $\endgroup$ – Quinn Culver Sep 22 '12 at 12:48
  • $\begingroup$ 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? $\endgroup$ – Math Girl Sep 22 '12 at 17:05
  • $\begingroup$ What about the Monotone Convergence Theorem? $\endgroup$ – Quinn Culver Sep 23 '12 at 14:43
  • $\begingroup$ 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 $f$) $\endgroup$ – Math Girl Sep 23 '12 at 16:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.