Let $f,g: [a,b] \rightarrow \mathbb{R}$ continuous functions such that $\int_a^{b} f(x)dx = \int_a^{b}g(x)dx$. Proof that there is $c \in [a,b]$ such that $f(c)=g(c).$

This questions has been asked before in this site, but the answer is wrong and I am studying real analysis so we must not to use the fundamental theorem of calculus, just upper and lower (Darboux) integrals.

My Attempt: I am trying this way: suppose that $f(c) \neq g(c), \hspace{0.2cm} \ \forall c \in [a,b], $ that is, suppose $f(c) - g(c)>0$ (whatever).

Since $f$ and $g$ are continuous so $f-g$ is continuous, that is, for every $\epsilon >0 $, there is an $\delta>0$ such that $x \in (c-\delta, c+ \delta)$ implies $f(x)-g(x)> 0. $ (Call $\epsilon= \frac{f(c) - g(c)}{2})$

Considerer the partition $P = \{x_0,x_1,...,x_n\}$ of interval $[a,b]$ such that $\{c - \delta, c+ \delta\} \subset P$ then there is $j\in \{1,2,..n-1\}$ such that $x_{j-1} = c-\delta$ and $x_j = c+\delta.$

we have:

$m_j(f-g) = inf \{f(x)-g(x) ; x \in [c-\delta,c+\delta] \}>0$.

Then the lower Darboux sum of ƒ with respect to P is: $L_{f,P} = \displaystyle \sum_{i=1}^{j-1}m_i(f-g)(x_i-x_{i-1}) + m_j(f-g)2\delta $ + $\displaystyle \sum_{i=j+1}^nm_i(f-g)(x_i-x_{i-1})$

and my objective is to show that $L_{f,P} $ is positive so I can conclude that the $\int_a^b f(x) - g(x)dx$ is positive and this is a contradiction but the only thing that I know is $m_j(f-g)2\delta>0.$

  • $\begingroup$ Do you know if $f-g$ attains its minimum and maximum values on $[a,b]$? $\endgroup$ – kimchi lover Mar 3 at 2:50
  • 2
    $\begingroup$ Since $f,g$ are continuous. if $f(c) \neq g(c)$ for all $c \in [a,b]$, then $f(x) - g(x) > 0$ or $f(x) - g(x) < 0$ for all $x \in [a,b]$. What would this say about the two integrals? $\endgroup$ – Ethan Alwaise Mar 3 at 2:53
  • $\begingroup$ did you use the intermediate value theorem, right? $\endgroup$ – user638057 Mar 3 at 3:07

If there is no $c \in [a, b]$ with

$f(c) = g(c), \tag 1$


$\forall x \in [a, b], \; f(x) - g(x) \ne 0; \tag 2$

this implies

$f(x) - g(x) > 0, \; \forall x \in [a, b] \tag 3$


$f(x) - g(x) < 0, \; \forall x \in [a, b]; \tag 4$

in the former case we have

$\displaystyle \int_a^b f(x) \; dx - \int_a^b g(x) \; dx = \int_a^b (f(x) - g(x)) \; dx > 0; \tag 5$

in the latter

$\displaystyle \int_a^b f(x) \; dx - \int_a^b g(x) \; dx = \int_a^b (f(x) - g(x)) \; dx < 0; \tag 6$

each of (5) and (6) imply

$\displaystyle \int_a^b f(x) \; dx \ne \int_a^b g(x) \; dx; \tag 7$

it follows then by contraposition that

$\displaystyle \int_a^b f(x) \; dx = \int_a^b g(x) \; dx \Longrightarrow \exists c \in [a, b], \; f(c) = g(c). \tag 8$

  • 1
    $\begingroup$ Thank you, I should use the intermediate value theorem to concluse that (1) implies (3) and (4) but i didn't remember of him. This way it's easier than my attempt. $\endgroup$ – user638057 Mar 3 at 3:11
  • $\begingroup$ @user638057: yes, the intermediate value theorem is ultimately what allows one to conclude that $f(x) \ne g(x), \forall x \in [a, b] \Longrightarrow $f(x) - g(x) <> 0$; I was thinking of mentioning it explicitly but it seemed pretty obvious so . . . $\endgroup$ – Robert Lewis Mar 3 at 3:17

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