Show that a function $f$ : $\mathbb{R}^d$$C$ is measurable if and only if it is the pointwise almost everywhere limit of continuous functions $f_n$ : $\mathbb{R}^d$$\mathbb{C}$.

I have proven the forwards direction, supposing $f_n$ : $\mathbb{R}^d$$\mathbb{C}$ has pointwise a.e. limit $f$ then it is measurable. I need help in the backwards direction.

To prove the backwards direction: So far, I think I have to use Egorov's theorem. So there exists a Lebesgue measureable set A of measurable set A of measure at most some $\epsilon$>0 such that $f_n$ converges locally uniformly to f on $\mathbb{R}^d$\A.

  • $\begingroup$ Egoroff's theorem is only valid for spaces of finite measure. $\endgroup$
    – abhi01nat
    Oct 21, 2020 at 5:12
  • $\begingroup$ Any suggestions on how I should begin the proof? $\endgroup$
    – Frances
    Oct 21, 2020 at 5:18

1 Answer 1


I will use the following well known result: If $g$ is integrable on $\mathbb R^{d}$ then, for any $\epsilon >0$, we can find a continuous function $h$ such that $\int |g-h| <\epsilon$.

For each $n$ there exists a continuous function $\phi_n$ such that $\phi_n(x)=1$ if $\|x||<n$ and $0$ if $\|x|| >n+1$.

Consider $\phi_n \arctan f$. This function is integrable. Hence there exists continuous function $f_n$ such that $\int |\phi_n \arctan f -f_n| <\frac 1n$. This implies that $\phi_{n_k} \arctan f -f_{n_k} \to 0$ almost everywhere for sum subsequence ${n_k}$. It follows that $ \arctan f -f_{n_k} \to 0$ almost everywhere (since $\phi_{n_k}(x)=1$ fior $\|x\|\leq n_k$). Now $\tan f_{n_k} \to f$ almost everywhere.


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