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.

Answer 1

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

Answer 2

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 $b \rightarrow \infty$.

Answer 3

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 $$\displaystyle \lim_{n \to \infty} \int_n^\infty f(x) \, dx = \lim_{n \to \infty} \int_{\mathbb R} f_n(x) \, dx = \int_{\mathbb R} \lim_{n \to \infty} f_n(x) \, dx = \int_{\mathbb R} 0 = 0$$