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Let $f : [a,+\infty[ \rightarrow \mathbb{R}_+$ a positive function, locally Riemann-integrable. It's assumed the improper integral $\int_{a}^{+\infty} f(x) dx$ is convergent.

For all $n \in \mathbb{N}$, we have $f_n = 1_{[a,n]} f$.

With the monotone convergence theoreme, we would like to prove that $f$ is Lebesgue-integrable on $[a,+\infty[$ and that $\int_{[a,+\infty[} f d\lambda(x) = \int_a^{+\infty} f(x) dx$.

I can constate that $\lim_{n \rightarrow +\infty} f_n = f$. Then, $\int_{[a,+\infty[} f d \lambda = \lim_{n \rightarrow +\infty} \int_{[a,n]} f_n d\lambda$.

For me, it is obvious that $\int_{[a,n]} |f_n| d \lambda = \int |1_{[a,n]} f| d \lambda < +\infty$ but I can't prove it. Could someone help me ?

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Note that $f_n(x)=f(x) 1_{[a,n]} \to f(x) 1_{[a,+\infty]}=F(x)$ pointwise.

Thus using the triangle inequality we have that $$||f_n(x)|-| F(x) || \leq |f_n(x) -F(x)| \to 0$$

Thus $|f_n(x)| \to |F(x)|$ pointwise.

Also $|F(x)|$ is Lebesgue integrable and $|f_n| \leq F(x),\forall n \in \Bbb{N}$

Now use the Dominated Convergence Theorem.

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