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I know this might be a duplicate, but I just wanted to see if my proof is correct.

I want to prove that if $f,g: [a,b]\rightarrow \mathbb{R}$ is bounded and Riemann integrable then $\max\{f,g\}$ and $\min\{f,g\}$ will also be Riemann integrable on $[a,b]$.

In order to show this, I've decided to prove that if $f$ is RI, then $|f|$ is RI. And then by arithmetic of limits, $$\max(f,g)(x) = \frac{f(x)+g(x) + |f(x)-g(x)|}{2}$$ and $$\min(f,g)(x) = \frac{f(x)+g(x) - |f(x)-g(x)|}{2}$$ will also be RI. My proof is as follows:

Since $f,g$ are RI on $[a,b]$, then by arithmetic of limits we know that $f+g$ is RI on $[a,b]$ and similarly $f-g$ is RI on $[a,b]$** (do I need to prove this or can this be safely assumed from the fact that $f-g = f+(-g)$?. And so, all there is left to show is that if $f$ is RI, then $|f|$ is RI.

By triangle inequality, we have that:

$\begin{align}\sup_{x \in [x_{j-1},x_j]}|f|(x) - \inf_{x \in [x_{j-1},x_j]}|f|(x) \\&=\sup_{x,y \in [x_{j-1},x_j]}| |f(x)|-|f(y)|| \\ &\leqslant \sup_{x,y \in [x_{j-1},x_j]}| f(x)-f(y)| \\ & \leq \sup_{x \in [x_{j-1},x_j]}f(x) - \inf_{x \in [x_{j-1},x_j]}f(x).\end{align}$

And thus, there exists a partition $P$ such that:

$U(P,|f|) - L(P,|f|) = \sum_{j=1}^n \left(\sup_{x \in [x_{j-1},x_j]}|f|(x) - \inf_{x \in [x_{j-1},x_j]}|f|(x)\right)(x_j - x_{j-1})\\ \leqslant \sum_{j=1}^n \left(\sup_{x \in [x_{j-1},x_j]}f(x) - \inf_{x \in [x_{j-1},x_j]}f(x)\right)(x_j - x_{j-1})\\= U(P,f) - L(P,f)$

Since $f$ is RI, we have that:

$U(P, |f|) - L(P, |f|) \leqslant U(P,f)-L(P,f) < \epsilon.$

Therefore, by the arithmetic of RI functions and the fact that if $f$ is RI then $|f|$ is RI, we have that $\max\{f,g\}$ and $\min\{f,g\}$ are RI.

Is this a sufficient proof? Or am I oversimplifying things/do I have to do a direct $\epsilon$ proof?

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Nikitau
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  • The inequality of the suprema is not true. Let $g(x)=1$ and $f(x)=-x^3$, now choose the interval $I:=[0,2]$. Observe that $\sup |f(I)|-\inf|g(I)|=8-1=7$, but $\sup f(I)-\inf(I)=0-1=-1$. – Masacroso Dec 12 '16 at 02:55
  • Sorry that was a typo. It should read $\leq \sup f(x) - \inf f(x)$. I understand the idea behind this is that we are finding a larger because we are taking the difference between something very large (sup f(x)) and something very small (inf f(x)). – Nikitau Dec 12 '16 at 03:01

2 Answers2

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Why not prove this directly with the Riemann criterion.

For example, in the first case, $h = \max\{f,g\}$, consider a partition $P = (x_0,x_1, \ldots, x_n)$ of $[a,b].$ With $I_k = [x_{k-1},x_k]$ define $M_k(h) = \sup_{x \in I_k}h(x)$ and $m_k(h) = \inf_{x \in I_k}h(x),$ with similar notational definitions for $M_k(f)$, $M_k(g)$, etc.

We have for all $x \in I_k$,

$$h(x) = \max[f(x),g(x)] \leqslant \max[M_k(f),M_k(g)] \\ \implies M_k(h) = \sup_{x \in I_k}h(x) \leqslant \max[M_k(f),M_k(g)], $$

and

$$h(x) = \max[f(x),g(x)] \geqslant f(x), \, g(x) \implies \\ m_k(h) = \inf_{x \in I_k}h(x) \geqslant \inf_{x \in I_k}f(x) = m_k(f) \\ m_k(h) = \inf_{x \in I_k}h(x) \geqslant \inf_{x \in I_k}g(x) = m_k(g) $$

Thus,

$$M_k(h) - m_k(h) \leqslant \max[M_k(f) - m_k(f), M_k(g) - m_k(g)] \\ \leqslant M_k(f) - m_k(f) + M_k(g) - m_k(g). $$

Forming upper and lower sums, it follows that

$$U(P,h) - L(P,h) \leqslant U(P,f) - L(P,f) + U(P,g) - L(P,g),$$

and applying the Riemann criterion given that $f$ and $g$ are Riemann integrable , we can find a partition $P_\epsilon$ for any $\epsilon > 0$ such that $U(P_\epsilon,h) - L(P_\epsilon,h) < \epsilon.$

RRL
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  • Wow. I'm feeling rather silly. This method is definitely a lot eaiser and more straight forward. Thanks! – Nikitau Dec 12 '16 at 07:12
  • You're welcome. Using $\max(f,g) = [f +g + |f - g|]/2$ is OK but tedious if you have to justify all the intermediate results. – RRL Dec 12 '16 at 07:18
  • I actually don't get the final bit. applying the Riemann criterion given that $f$ and $g$ are Riemann integrable , we can find a partition for $f$ and a partition for g such that $U(P_1,g) - L(P_1,g)< \epsilon$ and $U(P_2,f) - L(P_2,f)< \epsilon$, but how can you guarantee that you can find $P_1=P_2$ @RRL – H44S Nov 29 '19 at 20:49
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    @H44S: As you say find $P_1,P_2$ such that $U(P_1,g) - L(P_1,g)< \epsilon/2$ and $U(P_2,f) - L(P_2,f)< \epsilon/2$. Taking a common refinement $P_\epsilon = P_1\cup P_2$ we have $L(P_1,g) \leqslant L(P_\epsilon,g) \leqslant U(P_\epsilon,g) \leqslant U(P_2,g)$ and thus $U(P_\epsilon,g) -L(P_\epsilon,g) \leqslant U(P_1,g) - L(P_1,g) < \epsilon/2$. Similarly we get $U(P_\epsilon,f) -L(P_\epsilon,f) < \epsilon/2$. – RRL Nov 29 '19 at 21:07
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I should leave this in the comment but I do not have enough reputation point, so sorry for bothering.

I agree with RRL's answer but I am confused at this step:

Thus,(ℎ)−(ℎ)⩽max[()−(),()−()]⩽()−()+()−().

Why don't we write down $M_k(h) - m_k(h)\le max[M_k(f) - m_k(f),M_k(f) - m_k(g),M_k(g) - m_k(f),M_k(g) - m_k(g)]$

And all others remain the same. The following steps work out exactly in the same way, we will have $M_k(h) - m_k(h)\le2*(M_kf-m_kf+M_kg-m_kg)$ then.

ZKK
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