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I must construct a function $f: [0,\infty) \to \mathbb{R}$, but not Lebesgue-Integrable, but $f$ in $[0,c)$ is Riemann-integrable and $$\lim_{c \to \infty} \int_{0}^{c} f(x)dx$$ exists.

What I am thinking is: $f(x)=1/x$ on $\mathbb{R}$ and $f(x)=c, x=0$, which is measurable but it is not Lebesgue-Integrable because both sides of Lebesgue integral positive and negative must be finite.

Problem is that limit of this function doesn't exist since it is $\infty$, so it diverges and so it is not Riemann-Integrable.

What should I do? Thanks in advance!

gt6989b
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Melina
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  • The title is misleading: A Riemann integrable function is always Lebesgue integrable. You're talking about an improper Riemann integral. – zhw. Nov 12 '15 at 21:16

1 Answers1

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You can show that : $t\mapsto\frac{\sin t}{t}$ on $[0,+\infty[$ is "Riemann-integrable" (more precisely it's a generalized integral, i.e., limits of Riemann integrals) but that $t\mapsto\frac{\sin t}{t}$ is not Lebesgue-integrable on $[0,+\infty[$ (because $t\mapsto\left\lvert\frac{\sin t}{t}\right\rvert$ is not integrable on $[0,+\infty[$ and so the Lebesgue integral can't be defined).

Note that we talk about "Riemann-integrable" function on compacts intervals only, and that all the integrals of functions on (semi)-open intervals are get as Riemann-integrals limits, and that Riemann-integrable implies Lebesgue-integrable : indeed, for stairs functions $g$ and $h$ such as $g\leq f\leq h$, we have the equalities

$$\int_a^bg(x)\,dx=\int_{[a,b]}g\,d\lambda\leq\int_{[a,b]}f\,d\lambda\leq\int_{[a,b]}h\,d\lambda=\int_a^bh(x)\,dx$$ showing that Riemann-integrable implies Lebesgue-integrable in all generality.

davyjones
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Balloon
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    The subtle point is that the Lebesgue integral $\int_{0}^{\infty}f(x)dx$ is defined as $\int_Af(x)dx+\int_Bf(x)dx$ where $A={x\geq 0 :f(x)\geq 0}$ and $B={x\geq 0 :f(x)<0}$ which is not the same as $\lim_{c\to \infty}\int_0^c f(x)dx$. – DanielWainfleet Nov 12 '15 at 21:21