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I am trying to prove that If $g_n$ is a sequence of increasing functions on $[a,b]$ which converges uniformly to $g$, and if an increasing function $f$ is integrable w.r.t $g_n$ for all $n$, then f is integrable with respect to $g$ and $$\int_a^b f dg=\lim \int_a^b fdg_n$$ Can anyone give me a hint?

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Using integration by parts,

$$\int_a^b f \, dg_n = f(b)g_n(b) - f(a)g_n(a) - \int_a^b g_n \, df.$$

Hence,

$$\lim_{n \to \infty}\int_a^b f \, dg_n= \lim_{n \to \infty}[f(b)g_n(b) - f(a)g_n(a)]- \lim_{n \to \infty}\int_a^b g_n \, df \\ = f(b)g(b) - f(a)g(a)- \lim_{n \to \infty}\int_a^b g_n \, df .$$

Since $g_n$ converges uniformly to $g$ we can interchange limit and integral as

$$\lim_{n \to \infty}\int_a^b g_n df = \int_a^b g \, df. $$

The interchange is valid as

$$\left|\int_a^b g_n df - \int_a^b g \, df \right| = \int_a^b (g_n - g) df \leqslant \sup_{x \in [a,b]}|g_n(x) - g(x)|[f(b) - f(a)] \to 0.$$

Using a second integration by parts,

$$\lim_{n \to \infty}\int_a^b f \, dg_n= f(b)g(b) - f(a)g(a)-\int_a^b g_ \, df = \int_a^bf \, dg.$$

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