Let $A$ be a bounded, measurable susbset of $\mathbb{R}$. Prove that if $(f_n) \subset L^2 (A)$ converges uniformly to $f$ on $A$, then $f\in L^2(A)$ and $\lim_{n\rightarrow\infty} \int_A f_n = \int_Af$.
Clearly $f$ is measurable. We have to show that $\int_A f^2 < \infty$. Choose $\epsilon>0$. From uniform convergence we can find $N\in\mathbb{N}$ such that $|f_{n_0}(x)-f(x)|<\epsilon$ for all $x\in A$ and some $n_0>N$. Now denote by $A^-$ the part of $A$ where $f$ is negative and by $A^+$ the part of $A$ where $f$ is positive. Since $f$ is measurable, both sets are measurable.
We have $-f(x)<\epsilon - f_{n_0}(x)$ thus $\int_{A^-} f^2< \int_{A^{-}}(\epsilon-f_{n_0})^2<\infty$ (since A is bounded).
Similarly, since $f(x)<\epsilon + f_{n_0}(x)$, we get $\int_{A^+} f^2< \int_{A^{+}}(\epsilon+f_{n_0})^2<\infty$ and we can conclude that $f\in L^2 (A)$.
It remains to show that $\lim_{n\rightarrow\infty} \int_A f_n = \int_Af$. First we establish convergence in $L^2$ on $A$:
$$\|f_n-f\|_2^2=\int_A |f_n(x)-f(x)|^2\,dx \leq m(A) \left(\sup_{x \in A}|f_n(x)-f(x)|\right)^2 \to 0$$
Now we can use CS inequality and conclude that $\lim_{n\rightarrow\infty} \int_A f_n = \int_Af$.
Is my derivation correct?
No comments:
Post a Comment