If $f(x)$ is Integrable, $\int_{0}^{\infty}f(x)dx$ converges and $g(x) \geq x \Rightarrow \lim_{x \to \infty} \int_{x}^{g(x)}f(t)dt = 0$

Question

Prove or disprove:

If $f(x)$ is Integrable at $[0,t], \forall t > 0$ and the integral: $\int_{0}^{\infty}f(x)dx$ converges and $g(x) \geq x, \forall x \geq 0$ therefore $\lim_{x \to \infty} \int_{x}^{g(x)}f(t)dt = 0$

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What i did was to use Chushy, saying that:

$$ \forall \varepsilon > 0, \exists N \in N, \forall s \geq r>N: |\int_{r}^{s}f(x)dx| < \varepsilon $$

And this is exactly the term for limit of a function by cushy, so i conlcluded:

$$ \lim_{s,r \to \infty} \int_{r}^{s}f(x)dx = 0 $$

And specificaly we can take:

$$ g(x) = s, x = r \Rightarrow g(x) \geq x > N $$

Therefore getting:

$$ \lim_{x \to \infty}\int_{x}^{g(x)}f(t)dt = 0 $$

Therefore the statement is true.

A friend of mine asked me how could i do the translation from $f(x)$ to $f(t)$, i mean changing from $x$ to $t$ and i didnt realy know what to tell him.

And do you think my proof is correct?

Answer 1

Let $$F(X)=\int_0^Xf(t)dt$$ for $X\ge 0$.

$\int_0^\infty f(t)dt $ converges $ \iff $ $\lim_{X\to +\infty}F(X)=L\in \Bbb R$

on the other hand, $g(x)\ge x$ for any $ x\ge 0$, thus

$$\lim_{x\to +\infty}g(x)=+\infty$$

By composition of limits, we conclude that $$\lim_{x\to+\infty}\int_x^{g(x)}f(t)dt=$$

$$\lim_{x\to +\infty}(F(g(x))-F(x))=L-L=0$$