Here is the problem: Suppose that $\{f_n\}$ is a sequence of measurable functions satisfying $|f_n(x)| < M$ for all $n$ and all $x\in[0,1]$ and suppose that this sequence converges pointwise to the function $f$. Prove that the sequence converges in the mean to $f$.

Here is my attempted proof: By pointwise convergence of $f_n$ we have that $\lim |f_n-f| = 0$. The Bounded Convergence Theorem states that the integral of the limit equals the limit of the integral, i.e. $$\lim \int f_n \, d\mu = \int \lim f_n \, d\mu = \int f \, d\mu.$$ It then follows that $$\lim \int |f_n-f| \, d\mu = \int \lim |f_n-f| \, d\mu = \int 0 \,d\mu = 0.$$

I feel like there is something wrong here, but I am not sure what. Thoughts?

  • $\begingroup$ Is $f$ a measurable function? $\endgroup$ – GSF May 1 '17 at 22:17
  • $\begingroup$ Yes, I think so. $\endgroup$ – Hugh Mungus May 1 '17 at 22:18

First of all we have that $\{f_n\}_n\in L^1[0,1]$.

Moreover, being $|f_n(x)|<M$ for all $n$ and for all $x\in[0,1]$ the sequence converges (by DCT) in $L^1[0,1]$ to some $\tilde f\in L^1[0,1]$.

Note that $L^1[0,1]$ convergence implies a.e. pointwise convergence, up to passing to a subsequence, i.e. there exists $\{n_k\}_k$ such that $f_{n_k}\to\tilde f$ a.e. on $[0,1]$.

But by hypotesis $f_n$ and thus $f_{n_k}$ converges a.e. on $[0,1]$ to $f$ and by uniqueness of limits we have that $\tilde f=f$ a.e. on $[0,1]$.


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