Question about applying Dominated Convergence Theorem

Question

Question:

$\phi_n(x)=\int_{x_0}^xf(t,\phi_n(t))dt$, where $\phi_n(x)$ is continuous on $[a,b]$ and $f$ is continuous and bounded on $[a,b]\times(-\infty,+\infty)$.

1. If $\phi_n(x)$ converges uniformly to $\phi(x)$, prove that $\phi(x)=\int_{x_0}^xf(t,\phi(t))dt$.

2. If $\phi_n(x)$ just converges to $\phi(x)$, can we draw the same conclusion by applying Dominated Convergence Theorem like this:

$|f|\leqslant M,\lim\limits_{n\to\infty}\int_{x_0}^xf(t,\phi_n(t))dt=\int_{x_0}^x\lim\limits_{n\to\infty}f(t,\phi_n(t))dt=\int_{x_0}^xf(t,\phi(t))dt$, the first equation by DCT and the second by continuity of $f$.

Attempt:

1. Since $\phi_n$ uniformly converges to $\phi$,then $\phi_n$ is bounded and $\phi$ is continuous on $[a,b]$. So $f$ is uniformly continuous on $[a,b]\times[-M,M]$,where $|\phi_n|,|\phi|<M$.

Therefore $f(t,\phi_n(t))$ converges uniformly, so we can exchange the limit and integral.

Is that correct? Thanks for checking!

2. I'm wondering if DCT is correctly used here. Can we use $M(|f|<M)$ as the dominating function?

Answer 1

DCT does work here(in the sense of Lesbegue integral),for $g(t)=f(t,\phi(t))$ is measurable(as the limit of a continuous function series) and integrable($|g|\leq M$ on $[a,b]\times [-R,R]$,for some $R$ large enough).However,$g(t)$ may not be Riemann integrable since $\phi(t)$ may not be integrable.