Riemann and Lebesgue Integral, a simple question.

Question

Let $I=(0,1]$. Consider the measure space $(I,\mathcal{L}(I),\lambda)$, where $\lambda$ is the Lebesgue measure on $I$. I have just proved that $$\int_0^{\infty}\bigg|\frac{\sin x}{x}\bigg|\;dx=\infty.$$

From this how do I deduce that $$\int_\mathbb{{R}_+}\bigg|\frac{\sin x}{x}\bigg|\;d\lambda=\infty.$$

Thanks!

Answer 1

First I guess that $\lambda$ is the Lebesgue measure on $\mathbb R_+$, not just on $I=(0,1]$, otherwise the integral $\int_\mathbb{{R}_+}\big|\frac{\sin x}{x}\big|\;d\lambda$ is not defined.

Generally speaking, any Riemann-integrable function on an interval $[a,b]$ is also Lebesgue-integrable and the two integrals are equal: so $$ \int_0^{b}\bigg|\frac{\sin x}{x}\bigg|\;dx=\int_{[0,b]}\bigg|\frac{\sin x}{x}\bigg|\;d\lambda $$ for any $b$. The improper integral $\int_0^{\infty}\big|\frac{\sin x}{x}\big|\;dx $ in the Riemann sense is defined by $$ \int_0^{\infty}\bigg|\frac{\sin x}{x}\bigg|\;dx=\lim_{n\to\infty}\int_0^{n}\bigg|\frac{\sin x}{x}\bigg|\;dx, $$ (we can use the limit over integers $n$, since the integrals over $[0,b]$ are monotonically increasing with $b$). If you define the functions $f_n$ by $$ f_n(x):=\bigg|\frac{\sin x}{x}\bigg| \textrm{ if } x\in [0,n],\quad f_n(x):=0 \textrm{ if } x>n $$ then, as Lebesgue integrals, $$ \int_{[0,n]}\bigg|\frac{\sin x}{x}\bigg|\;d\lambda=\int_{\mathbb R_+}f_n(x)\;d\lambda. $$ Now apply the monotone convergence theorem: since $f_n(x)\to \big|\frac{\sin x}{x}\big|$ pointwise and monotonically, you have $$ \int_\mathbb{{R}_+}\big|\frac{\sin x}{x}\big|\;d\lambda =\lim_{n\to \infty}\int_{\mathbb R_+}f_n(x)\;d\lambda=\lim_{n\to \infty}\int_0^{n}\bigg|\frac{\sin x}{x}\bigg|\;dx=\infty, $$ since you already know that the improper Riemann integral is $+\infty$.

Answer 2

The function $\frac{\sin(x)}{x}$ is defined on $[0;1]$, but as $\sin(x) \sim x$ close to $1$, we can extend this function on $[0;1]$ by a continuous function. From now, I identify this function obtained as a continuous extension of $\frac{\sin(x)}{x}$ with $\frac{\sin(x)}{x}$. As $|\frac{\sin(x)}{x}|$ is positive and bounded on $[0;1]$ (by what we have said above, and of course by continuity of the absolute value and the fact that a composition of continuous function is continuous), in particular it's the case on every compact of $[0;1]$, and then we have : $$ \int_{0}^{+\infty}|\frac{\sin(x)}{x}|dx < +\infty$$ if and only if $$\int_{\mathbb{R}_+}|\frac{\sin(x)}{x}|d\lambda < +\infty \\ \textrm{ and } \textrm{ the set of points of } [0;1] \textrm{ where the function } |\frac{\sin(x)}{x}| \textrm{ is not continuous has measure } 0 $$

From what we said below, you find that we must have $\int_{\mathbb{R}_+}|\frac{\sin(x)}{x}|d\lambda = +\infty$ (cause of course $\lambda(\{0\}) = 0$).