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Let $f$ be a function that is Lebesgue integrable (w.r.t. the Borel measure $\lambda$) over $(0, \infty)$. Let $a, x > 0$ with $a < x$ and let $(a_n)_{n \in \mathbb{N}}$ be an increasing sequence of positive numbers converging to $a$.

Under what conditions does the equality $$ \lim_{n \to \infty} \int_{(a_n, x)} f(t) \text{d}\lambda(t) = \int_{(a, x)} f(t) \text{d}\lambda(t) $$ hold?

I know a lot of theorems about limits coupled with integrals, but only if the limit-variable ($n$ in our case) appears in the integrand, but here, it appears in the set over which we integrate (the boundaries of integration).

Note that $f$ is not (necessarily) Riemann integrable, because in that case, the Riemann and Lebesgue integral of $f$ (over compact sets) coincide, and if the integrals in the expression were Riemann integrals, we know that the equality holds.

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$\lvert f\cdot I_{(a_n,x)}\rvert\le\lvert f\cdot I_{(0,\infty)}\rvert$ and $\int \lvert f\cdot I_{(0,\infty)}\rvert<\infty$ (since you specified that $f$ is integrable over $(0,\infty)$), and your result follows from the Lebesgue dominated convergence theorem.

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  • $\begingroup$ How does this follow from the DCT? Note that the output of the integral is not a function, because $x$ and $a$ are fixed numbers in $\mathbb{R}$, and not variables. $\endgroup$ – limitIntegral314 Dec 18 '16 at 0:07
  • $\begingroup$ See the updated answer. $\endgroup$ – Momo Dec 18 '16 at 0:12
  • $\begingroup$ So the clue is that integrating $f$ over $(a_n, x)$ is the same as integrating $f \cdot I_{(a_n, x)}$ over (say) the set $(0, x)$ (or more generally, over any set containing $(a_n, x)$). Right? $\endgroup$ – limitIntegral314 Dec 18 '16 at 0:22
  • $\begingroup$ Yes, $\int_A f=\int f\cdot I_A$ $\endgroup$ – Momo Dec 18 '16 at 0:23

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