Let $g(x)\ge0$. If $\int_a^bg(x)dx=0$, show that $\int_a^bf(x)g(x)dx=0,$ where $f$ is any integrable function.

If simeone is allowed to use the Mean Value thorem for integrals, the proof is at hand. But for that $f$ must be continuous!

Any suggestion?

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    $\begingroup$ Just show that $g$ is identically zero. $\endgroup$ – Artem May 20 '19 at 1:06
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    $\begingroup$ If $g \ge 0$ and $\int ^b_a g(x) dx = 0$ then $g = 0$ almost everywhere, and so $fg =0$ almost everywhere as well. $\endgroup$ – User8128 May 20 '19 at 1:07
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    $\begingroup$ @Artem it isn't, necessarily $\endgroup$ – operatorerror May 20 '19 at 1:07
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    $\begingroup$ @qbert Ooof, I've been teaching introductory calculus too long. Silly me. Forgot. $\endgroup$ – The Count May 20 '19 at 2:47
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    $\begingroup$ @TheCount no sweat, I generally distrust the Riemann-integral myself :P $\endgroup$ – operatorerror May 20 '19 at 2:55

I try to prove this without invoking anything about Lebesgue integral. Moreover, I do not assume the fact that $fg$ is Riemann integrable nor the inequality $| \int_a^b f(x)g(x) dx | \leq \int_a^b |f(x)g(x)|dx$.

$f$ is Riemann integrable $\Rightarrow$ $f$ is bounded. Choose $M>0$ such that $|f(x)|\leq M$ for all $x\in[a,b]$. Let $\varepsilon>0$ be given. Choose $\delta>0$ such that for any partition $\mathbb{P}=\{x_{0},x_{1},\ldots,x_{n}\}$ of $[a,b]$ (with $a=x_{0}<x_{1}<\ldots<x_{n}=b$) and any $\xi_{i}\in[x_{i-1},x_{i}]$, if $||\mathbb{P}||<\delta$ (here, $||\mathbb{P}||=\max_{1\leq i\leq n}|x_{i}-x_{i-1}|$), then $$ \left|\sum_{i=1}^{n}g(\xi_{i})(x_{i}-x_{i-1})-\int_{a}^{b}g(x)dx\right|<\frac{\varepsilon}{M}. $$ That is, $$ \left|\sum_{i=1}^{n}g(\xi_{i})(x_{i}-x_{i-1})\right|<\frac{\varepsilon}{M}. $$

Now, let $\mathbb{P}=\{x_{0},x_{1},\ldots,x_{n}\}$ be an arbitrary partition of $[a,b]$, with $a=x_{0}<x_{1}<\ldots<x_{n}=b$, that satisfies $||\mathbb{P}||<\delta$. Let $\xi_{i}\in[x_{i-1},x_{i}]$ be arbitrary. Then \begin{eqnarray*} & & \left|\sum_{i=1}^{n}f(\xi_{i})g(\xi_{i})(x_{i}-x_{i-1})-0\right|\\ & \leq & \sum_{i=1}^{n}|f(\xi_{i})g(\xi_{i})|(x_{i}-x_{i-1})\\ & \leq & M\sum_{i=1}^{n}g(\xi_{i})(x_{i}-x_{i-1})\\ & < & M\cdot\frac{\varepsilon}{M}\\ & = & \varepsilon. \end{eqnarray*} This shows that the Riemann integral $\int_{a}^{b}f(x)g(x)dx$ exists and $\int_{a}^{b}f(x)g(x)dx=0$.

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    $\begingroup$ Props for a complete answer that doesn't invoke the Lebesgue integral. $\endgroup$ – Charles Hudgins May 20 '19 at 4:44
  • $\begingroup$ @Alex Ortiz Although it is true that if $f$ and $g$ are Riemann integrable functions on $[a,b]$, then $fg$ is also Riemann integrable on $[a,b]$, this fact is not an axiom and it requires a proof. I tried to assume as less as possible and avoid invoking this fact. $\endgroup$ – Danny Pak-Keung Chan May 21 '19 at 21:40

This is about Riemann integration, so $f,g$ are necessarily bounded.

We'll use that.

$$\int_a^b f g dx =\int_a^b \underset{\mbox{nonnegative}}{\underbrace{(f - \inf f)}} g dx+ (\inf f )\underset{=0}{\underbrace{\int_a^b g dx}} $$

so the problem is reduced to Riemann integrable nonnegative $f$. For such $f$

$$ 0\le \int_a^b g f dx \le (\sup f) \int_a^b g dx =0.$$


Define $$ \begin{align} p_n(x) &=\frac{|x-n|-|x-n-1|-|x+n|+|x+n+1|}2\\[6pt] &=\left\{\begin{array}{cl} -1&\text{if }x\lt-n-1\\ x+n&\text{if }-n-1\le x\lt-n\\ 0&\text{if }-n\le x\lt n\\ x-n&\text{if }n\le x\lt n+1\\ 1&\text{if }n+1\le x \end{array}\right. \end{align} $$ In particular, $p_n$ is continuous, so that $p_n(f)$ is integrable. Furthermore, $$ \begin{align} \sum_{k=0}^{n-1}p_k(x) &=\frac{|x+n|-|x-n|}2\\ &=\left\{\begin{array}{cl} -n&\text{if }x\lt-n\\ x&\text{if }-n\le x\lt n\\ n&\text{if }n\le x\\ \end{array}\right. \end{align} $$ Therefore, $$ \begin{align} \left|\int_a^bf(x)\,g(x)\,\mathrm{d}x\right| &=\left|\sum_{n=1}^\infty\int_a^bp_n(f(x))\,g(x)\,\mathrm{d}x\right|\\ &\le\sum_{n=1}^\infty\int_a^b|p_n(f(x))|\,g(x)\,\mathrm{d}x\\ &\le\sum_{n=1}^\infty\int_a^b\,g(x)\,\mathrm{d}x\\ &=\sum_{n=1}^\infty0\\[6pt] &=0 \end{align} $$ Note that if $f$ is finite, then the sum above is finite.

  • $\begingroup$ One should try to avoid using anything about Lebesgue integral. $\endgroup$ – Danny Pak-Keung Chan May 20 '19 at 3:44
  • $\begingroup$ @DannyPak-KeungChan Actually the OP didn't say that... $\endgroup$ – Bach May 20 '19 at 3:50
  • $\begingroup$ Implicitly in the tag... I agree that OP should state this explicitly. $\endgroup$ – Danny Pak-Keung Chan May 20 '19 at 3:52
  • $\begingroup$ This really doesn't have anything to do with Lebesgue integration, and I have modified the answer to make that more apparent. $\endgroup$ – robjohn May 20 '19 at 4:08
  • $\begingroup$ I didn't check that the convergence $\sum_1^N p_n(f)\to_N f$ is uniform, so that you know it is valid to change integral and summation. May be worth mentioning. $\endgroup$ – Alex Ortiz May 21 '19 at 17:48

Since $g(x)\ge0$ and $\int_{a}^bg(x)=0$, then for every $\epsilon>0$, $m(\{x:g(x)\ge\epsilon\})=0$ where $m$ denotes the Lebesgue measure on $\mathbb R^1$. Set $\int_a^b |f(x)|dx=M<\infty$ since $f(x)$ is integrable. \begin{align} \int_a^b |f(x)g(x)|dx&=\int_a^b |f(x)|g(x) \chi_{\{x:g(x)<\epsilon\}}dx+\int_a^b |f(x)|g(x)\chi_{\{x:g(x)\ge \epsilon\}}dx\\ &\le \epsilon\int_a^b |f(x)|dx\\ &=\epsilon M \end{align} Since $\epsilon$ can be arbitrary small, we conclude that $\int_a^bf(x)g(x)dx=0$.

Or you may use the idea above to prove that $g(x)=0$ a.e. To do this, note that we can decompose the set $\{x:g(x)>0\}$ as

$$\displaystyle{\{x:g(x)>0\}=\bigcup_{n=1}^\infty\{ x: g(x)\ge\frac 1n \}}$$

and note that

$$ m(\{x:g(x)>0\})\le\sum_{n=1}^\infty m(\{ x: g(x)\ge\frac 1n \})=0 .$$

We can conclude that $g(x)=0$ a.e. This will give you $\int_a^b f(x)g(x)dx=0.$

  • $\begingroup$ It seems that OP prohibits the use of anything related to Lebesgue integral. If we are allowed to do so, then it is one line. $f\geq 0$ and $\int f=0$ implied that $f=0$ a.e.. Then $fg=0$ a.e., so $\int fg=0$. $\endgroup$ – Danny Pak-Keung Chan May 20 '19 at 3:21
  • $\begingroup$ But you need to prove that $f\ge 0$ and $\int f=0$ implies $f=0$ a.e. :) $\endgroup$ – Bach May 20 '19 at 3:30
  • $\begingroup$ This is an obviously fact. Let $A=\{x\mid f(x)>0\}$. Then $A=\cup_{n=1}^\infty \{x\mid f(x)\geq \frac{1}{n}\}$. If $\mu(A)>0$, then $\mu(A_n)>0$ for some $n$. Then $\int f \geq \int_{A_n} f \geq \frac{1}{n}\mu(A_n)>0$. Here, $A_n =\{x\mid f(x)\geq \frac{1}{n}\}$. $\endgroup$ – Danny Pak-Keung Chan May 20 '19 at 3:42
  • $\begingroup$ @DannyPak-KeungChan Well, basically the same thing as I mentioned in the latter part of my post... $\endgroup$ – Bach May 20 '19 at 3:53

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