If $f$ is Riemann integrable and $f>0$ for $x\in[a,b]$ then $\int{f>0}$

Question

I've seen the proof and it seems much less trivial than what I had in mind. Can anyone point out to me what is wrong about the following proof:

Let P be some partition of $[a,b]$ such that: $$ P= \begin{Bmatrix} a = x_0 < x_1 < x_2 < · · · < x_{n−1} < x_n = b \end{Bmatrix}$$

$$\forall x\in[a,b]: f(x)>0 \Rightarrow 0<\min_{x\in[x_{i-1},x_i]}f(x)\Longrightarrow 0<\sum_{i=1}^{n} \min_{x\in[x_{i-1},x_i]}{f(x)} \cdot \Delta x_{i}$$ let $L(f, P)$ be the lower sum of partition P

$$0<\sum_{i=1}^{n} \min_{x\in[x_{i-1},x_i]}{f(x)} \cdot \Delta x_{i}=L(f,P)\le\sup_{P}{L(f,P')}=\int_{a}^{b}f(x)dx $$

hence $$\int_{a}^{b}f(x)dx > 0$$

You do not know if $f$ actually attains its minimum on any interval — Andrew Shedlock, Mar 28 '20 at 05:55
It might be easier to show that if $f \ge 0$ and $\int f = 0$ then $f = 0$. — copper.hat, Mar 28 '20 at 06:01
@copper.hat Those conditions would imply that $f = 0$ a.e., not necessarily everywhere. But is this possible to prove without any Lebesgue machinery? — , Mar 28 '20 at 07:27
@copper.hat I guess this other answer of yours addresses that - you don't need the full notion of Lebesgue measure or Lebesgue integration, but you do need the notions of measure zero and content zero. — , Mar 28 '20 at 07:34
Alternatively, it suffices to show that a Riemann integrable function is continuous somewhere. In fact, it must have infinitely many points of continuity — , Mar 28 '20 at 07:43

Just dropped in · Accepted Answer · 2020-03-28T21:08:39.807

To be clear: the statement is correct,but the proof you are demonstrating is not. There is a very, very slight point that contains a mistake. You are saying that since $f(x)>0$ for all $x$, it is $\min_{x\in[x_{i-1},x_i]}f(x)>0$. Why do you think this minimum exists? If it is a minimum, then fine, it will be equal to $f(\xi_i)$ for some $\xi_i\in[x_{i-1},x_i]$ therefore strictly positive. But this is not always the case! The following two examples illustrate why this is not always the case:

For example: There is a famous theorem that says $f$ is Riemann integrable if and only if the discontinuities of $f$ form a set of measure zero. Now we can have sets of measure zero that are very, very weird. For example the Cantor set. Denote by $C$ the classic cantor set and define a function $f:[0,1]\to(0,\infty)$ by $f(x)=1$ if $x\not\in C$ and we define $f$ however we want on $C$, but it must be strictly positive. We can define $f$ to be positive and satisfying for any $\delta>0$

$$\inf_{x\in[1/3-\delta,1/3+\delta]}f(x)=0$$ (Look at the wikipedia link to visualize the Cantor set. Can you think of a way to define such an $f$? remember, you can define $f(x)$ however you like on $C$ and you will still get a Riemann integrable function. The only restriction is that $f(x)>0$ for all $x$)

This function provides a counter example, since if we partition $[0,1]$ in subintervals, $1/3$ has to belong to a subinterval, therefore the infimum of $f$ in this subinterval will be $0$.

Edit: a much simpler counter example: set $f(x)=1$ everywhere on $[0,1]$ except $\{0.5+1/n: n\geq3\}$. For these points, set $f(0.5+1/n)=1/n$. Then the same thing happens at $0.5$ (and of course $f$ is positive, Riemann integrable).

Does your counter example mean that: if $f>0$ then not always $\int{f}$>0? (for a Riemann integrable function) — hen shar, Mar 28 '20 at 17:35
@henshar No! It is indeed true that if $f>0$ then it is $\int f>0$, but the proof you are presenting here has an error. My counter-examples simply point out that error. — Just dropped in, Mar 28 '20 at 17:38
I think it would help to clarify what your counterexample is a counterexample to. A casual reader would incorrectly presume that is is a counterexample to the question. — copper.hat, Mar 28 '20 at 21:02

If $f$ is Riemann integrable and $f>0$ for $x\in[a,b]$ then $\int{f>0}$

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