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As a completion of this question here Show that the function $f(x)g(x)$ is integrable. I do not know how to answer $(b)$ and $(c)$ below.

Let $A:=[a,b].$ Suppose that the function $f: A \rightarrow \mathbb{R}$ is continuous, $g: A \rightarrow \mathbb{R}$ is integrable and $g(x) \geq 0$ for almost all $x \in A.$

$(a)$ Show that the function $f(x)g(x)$ is integrable. (I got an answer for this in the previous question)

$(b)$ Show that there exists a point $p \in A$ such that $$\int_{A} f(x)g(x) dx = f(p) \int_{A}g(x) dx \quad \quad (1)$$

$(c)$ Is $(1)$ valid in the case $A = [a,b] \cup [c,d]$ if $[a,b] \cap [c,d] = \phi.$

My questions are:

1- What does question $(b)$ want to teach us?

2- What does question $(c)$ want to teach us?

3- How can I solve question $(b)$ and question $(c)$?

Could anyone help me in answering those questions, please?

1 Answers1

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Question b) teaches us that multiplying the integrand ($g(x)$) by a continuous function amounts to multiplying the integral ($\int_A g(x)dx$) by one of the values of the continuous function. Question c) warns us that the result is non longer true if $A$ consists of several segments.

For question b), you know that $f$ is bounded and that its supremum and infimum are reached, because $f$ is continuous on the segment $[a,b]$. With $m = \inf f$ and $M = \sup f$, and because g is non-negative you can write

$$ m = \frac{\int_A mg(x) dx}{\int_A g(x) dx} \leq \frac{\int_A f(x)g(x) dx}{\int_A g(x) dx} \leq \frac{\int_A Mg(x) dx}{\int_A g(x) dx} = M. $$

Now you can just apply the intermediate value theorem with $f$: there exists $p \in A$ such that

$$ \frac{\int_A f(x)g(x) dx}{\int_A g(x) dx} = f(p). $$

Of course, this proof only works if $\int_A g(x)dx > 0$, but if $\int_A g(x)dx = 0$ the result is immediate.

For question c), you can exhibit a counter-example. For instance take $A = [0,1] \bigcup [2,3]$, $g=1$ and $f|_{[0,1]} = -1$, $f|_{[2,3]} = 1$. All the hypotheses are verified but there is no way we can write $(1)$ because both $f(p)$ and the integral of $g$ are non zero, whereas the integral of $f$ is zero.

DodoDuQuercy
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  • what about $(c)$? –  Jun 19 '20 at 14:52
  • Also, for part $(a)$ in the link mentioned .... do I have to approximate the continuous function $f$ with simple functions or the answer that is given is fine? –  Jun 19 '20 at 14:54
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    I have completed my answer concerning question c). For question a) in the link, it suffices to show that $fg$ is bounded in absolute value by an integrable function. As explained in the post, we can do this because $f$ is bounded (continuous on a segment) by a certain $K \geq 0$ so $|fg| \leq K g$, with $Kg$ integrable because $g$ is. Note that you should also prove that the function $fg$ is measurable, but this simply because $g$ is measurable by definition of an integrable function, $f$ is measurable because it is continuous, and the product of two measurable function is measurable. – DodoDuQuercy Jun 19 '20 at 15:02