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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.

My questions are:

1-I know that the product of 2 Riemann integrable functions is again a Riemann integrable function by this question here The product of two Riemann integrable functions is integrable but the product of 2 Lebesgue integrable functions is not necessarily Lebesgue integrable (am I correct ?). What is the theorem that gives us the necessary conditions for the product of 2 Lebesgue integrable functions to be integrable?

2- I studied from " Real Analysis " by Royden and Fitzpatrick, fourth edition. But still, I do not know how to prove the above question, so could anyone help me in proving it, please?

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    You can approximate a continuous function to within $\varepsilon$ by a sum of characteristic functions and then use the integrability of $g$. – Integrand Jun 18 '20 at 20:57
  • If a function is Riemann integrable then it is definitely Lebesgue Integrable. – Dinesh Jun 18 '20 at 20:57
  • @Integrand could you please write the details of your idea in an answer? and tell me if you agree with the answer given below or we must approximate? and if we must approximate, why we must do this? .... thanks in advance –  Jun 18 '20 at 22:37
  • @lebesgue I do not see how your comment answers my questions above .... could you please explain more? –  Jun 18 '20 at 22:38

1 Answers1

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Since $f$ is continuous on $[a,b]$, we have $|f(x)| \leqslant M$ and $|f(x) g(x)| \leqslant Mg(x)$ a.e..

You should be able now to conclude that $fg$ is integrable. See Proposition 16 (the Integral Comparison Test) in Royden.

RRL
  • 90,707
  • What about my first question? –  Jun 18 '20 at 22:39
  • The functions $f(x) = g(x) = 1/\sqrt{x}$ are Lebesgue integrable on $(0,1]$, but $f(x)g(x) =1/x$ is not. So to answer "am I correct" -- yes you are. As to when is the product integrable -- that is a broad question. I know of no concise set of necessary and sufficient conditions if that is what you want. A more general sufficient condition is if $f$ is bounded and $g$ is integrable. The same proof used above shows that $fg$ is integrable. – RRL Jun 18 '20 at 23:01
  • So what about approximating the function $f$ ..... do we need to do this? –  Jun 19 '20 at 03:48
  • Sorry for this stupid question .... why the function $1/x$ is not Lebesgue integrable? –  Jun 19 '20 at 03:50
  • Are you referring to the suggestion to approximate the continuous function $f$ with simple functions? That is somewhat of an overkill. – RRL Jun 19 '20 at 03:50
  • Yes exactly I am referring to this. –  Jun 19 '20 at 03:51
  • @Smart20: What is your understanding of what it means for a function to be Lebesgue integrable? I'm trying to see where you are having trouble here. Are you comfortable with if $g$ is integrable, then so is $Mg$ where $M$ is a constant.? – RRL Jun 19 '20 at 03:53
  • Yes I am comfortable with this –  Jun 19 '20 at 03:54
  • the proof not yet. –  Jun 19 '20 at 03:57
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    Read this answer. It summarizes the approach to defining the Lebesgue integral that Royden uses. Then you will see why $|fg| \leqslant Mg$ implies that $fg$ is integrable. If $h$ is any bounded measurable function on $[a,b]$ such that $h \leqslant |fg|$ then since $Mg$ is integrable we have $\int_{[a,b]} h \leqslant M \int_{[a,b]} g$. Thus $ \int_{[a,b]} |fg| := \sup_{h}\int_{[a,b]}h < \infty$. – RRL Jun 19 '20 at 04:08
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    So $|fg|$ integrable implies $fg$ integrable. Finally you must know that $\int_0^1 x^{-1} , dx = \infty$ As an improper integral and Lebesgue integral. – RRL Jun 19 '20 at 04:13
  • The key part here seems to be (I am no expert in Lebesgue stuff) that the integral needs to be a finite real number and thus bounded. If $f$ is unbounded then there is no guarantee that integral of $fg$ will remain bounded. – Paramanand Singh Jun 19 '20 at 06:56