Let $f \in L^1$ then prove $\lim_{b \rightarrow \infty} \int_b^{\infty} f(x) dx=0$.

Question

So the question is as stated in the title. We are given the hint to use LDCT.

Since this is homework I'm not looking for an explicit solution. I just need hints.

For example, my first thoughts were to rewrite $f$ as,

$$ f(x) = \lim_{n \rightarrow \infty} \sum_{i = 1}^n f_i(x) $$

Then to use the linearity of the integral,

$$ \int_b^{\infty}f(x)dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^n \int_b^{\infty}f_i(x)dx $$

I'm not sure how to show that the integral must go to zero as $b \rightarrow \infty$. I thought I would see a connection from the integrability of each $f_i$ and that somehow if each term didn't start to go to zero then we would reach a contradiction. But that doesn't seem right, now.

Consider $g := f \cdot \chi_{[b, \infty]}$ and use $\vert f \vert$ as your dominating function. $g$ converges to zero almost everywhere, use dominated convergence theorem. — Robert Cardona, Nov 15 '14 at 04:23
Thank you! I was trying forever to create the correct dominating function. I kept trying to find something just larger than $|f|$. — IQ472, Nov 15 '14 at 04:38

score 5 · Answer 1 · answered Nov 15 '14 at 04:42

Let $f \in L^1(\mathbb R)$, then we know $\int_{\mathbb R} \vert f(x) \vert \, dx < \infty$.

Define $f_n(x) = f(x) \cdot \chi_{[n, \infty]}$. Observe that $\vert f_n(x) \vert \leq \vert f(x)\vert$ for all $x \in \mathbb R$, where $\vert f \vert \in L^1(\mathbb R)$.

Notice that $\displaystyle \lim_{n \to \infty} f_n(x) = 0$ almost everywhere.

By the Lebesgue Dominated Convergence Theorem, we have $$\displaystyle \lim_{n \to \infty} \int_n^\infty f(x) \, dx = \lim_{n \to \infty} \int_{\mathbb R} f_n(x) \, dx = \int_{\mathbb R} \lim_{n \to \infty} f_n(x) \, dx = \int_{\mathbb R} 0 = 0$$

score 1 · Accepted Answer · answered Nov 15 '14 at 04:25

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HINT: $|f|$ must be integrable, hence for every $\varepsilon>0$ there exists $b_\varepsilon$ such that $$ \int_{b_\varepsilon}^{\infty} |f(x)|\, dx<\varepsilon. $$

answered Nov 15 '14 at 04:25

Przemysław Scherwentke

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First, I'm not sure I have heard of this before. Is this related to the absolute continuity of the integral? Or is this a property of the integrable function? – IQ472 Nov 15 '14 at 04:30
So by Cardona's hint, I can say: $f\chi_{[b,\infty]} \leq |f|$, thus, $\int_{b_{\epsilon}}^{\infty} |f(x)| dx < \epsilon$ – IQ472 Nov 15 '14 at 04:31
@IQ472 It is one of the definitions of convergence of an indefinite integral. – Przemysław Scherwentke Nov 15 '14 at 04:32
I totally see that, as it is similar to a convergent series! Thanks so much – IQ472 Nov 15 '14 at 04:35
@IQ472 And to your second comment: yes. – Przemysław Scherwentke Nov 15 '14 at 04:36
@IQ472 BTW: Congratulations! Estimated IQ of Terrence Tao is between 220 and 230. ;-) – Przemysław Scherwentke Nov 15 '14 at 04:42

score 1 · Answer 3 · answered Nov 15 '14 at 04:26

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Consider $f_b(x) = f(x) \textrm{ if } x \geq b$ and $f_b(x) = 0 \textrm{ otherwise}$.

Then $f_b$ is dominated by $f$ and $f_b \rightarrow 0$ as $b \rightarrow \infty$.

answered Nov 15 '14 at 04:26

Empiricist

7,933

Let $f \in L^1$ then prove $\lim_{b \rightarrow \infty} \int_b^{\infty} f(x) dx=0$.

3 Answers3