Integration inequality proof validation

[bonfire09]

Homework Statement

Let $f:[a,b]\rightarrow\mathbb{R}$ and $g:[a,b]\rightarrow\mathbb{R}$ be continuous functions having the property $f(x)\leq g(x)$ for all $x\in[a,b]$. Prove $\int_a^b \mathrm f <\int_a^b\mathrm g$ iff there exists a point $x_0$ in $[a,b]$ at which $f(x_0)<g(x_0)$. I just needed help with the reverse direction.

Homework Equations

$R(f,P_n)$ stands for the Riemann integral.

The Attempt at a Solution

Suppose there exists a point $x_0$ in $[a,b]$ such that $f(x_0)<g(x_0)$. Let $P_n$ be a regular partition for $[a,b]$ such that $\lim_{n\to\infty} ||P_n||=0$. Then there exists an interval $[x_{j-1},x_j]$ such that $x_0\in[x_{j-1},x_j]$. Let $c_j=x_0$ for some tag $c_j$ where $1\leq j\leq n$. Picking the same tags for both functions we have $R(f,P_n)=\sum_{i=1}^{n} f(c_i)(x_i-x_{i-1})=f(c_1)(x_1-x_0)+...+ f(c_j)(x_j-x_{j-1})+...+f(c_n)(x_n-x_{n-1})$ and $R(g,P_n)= \sum_{i=1}^{n} g(c_i)(x_i-x_{i-1})=g(c_1)(x_1-x_0)+...+g(c_j)(x_j-x_{j-1})+..._g(c_n)(x_n-x_{n-1})$. We see by assumption that $f(c_j)(x_j-x_{j-1})< g(c_j)(x_j-x_{j-1})$. Then $f(c_1)(x_1-x_0)+...+ f(c_j)(x_j-x_{j-1})+...+f(c_n)(x_n-x_{n-1})< g(c_1)(x_1-x_0)+...+g(c_j)(x_j-x_{j-1})+..._g(c_n)(x_n-x_{n-1})$. Thus $R(f,P_n)=\sum_{i=1}^{n} f(c_i)(x_i-x_{i-1})<R(g,P_n)= \sum_{i=1}^{n} g(c_i)(x_i-x_{i-1}) \implies R(f,P_n)<R(g,P_n)$. Since $f$ and $g$ are continuous that means they are integrable. Thus we have $\int_a^b \mathrm f =\lim_{n\to\infty} R(f,P_n)<\lim_{n\to\infty} R(g,P_n)=\int_a^b\mathrm g$

Here is my solution. I need to know if its right or not and how I can clean it up. Thanks.

 

[mfb]

I'm not sure if you have to make it so complicated. In addition, I don't see where you used the properties of continuous to show the inequality for the Riemann sums (you need it!).
Do you know some basic properties about integrals, like the integral of f-g if you know the individual integrals?

(f-g)(x₀) < 0, therefore there exists an interval around x₀ where ...

 

[bonfire09]

I know about those properties but I thought I would not need those. I thought that since f and g are continuous functions then their both integrable which is what I used to get my last step. Despite my proof being long would it suffice?

 

[mfb]

Hmm, I think I understand what you did - picking c_j is possible as f and g are continuous, so the justification of this steps comes a bit late. Okay, looks fine, just a bit complicated.

 

[pasmith]

I know about those properties but I thought I would not need those. I thought that since f and g are continuous functions then their both integrable which is what I used to get my last step. Despite my proof being long would it suffice?

The problem with your proof is that it obscures a key difference between integrable functions in general and continuous functions in particular. To that extent it doesn't promote understanding.

It is true for all integrable functions that if $f(x) < g(x)$ everywhere in $(a,b)$ then
$$\int_a^b f(x)\,dx < \int_a^b g(x)\,dx.$$

But if $f$ and $g$ are integrable and such that $f(x) = g(x)$ except at a finite number of points, then $$\int_a^b f(x)\,dx = \int_a^b g(x)\,dx.$$
So the restriction to continuous $f$ and $g$ must somehow exclude this case, and that's what you should focus on.

Also, proofs involving Riemann sums take considerable time to write out in examinations, so shorter proofs using basic properties of the integral are preferable where possible. (If you want practice in manipulating upper and lower sums, try proving that the second fact I asserted is indeed true.)