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I have the following problem:

Let $f,f_1,...,f_n,...\in L^1(\Omega)$. We assume that there exists $g\in L^1(\Omega)$ such that $|f_n|\leq g$ for all n. Show that $$f_n\stackrel{\mu}{\rightarrow}f\Rightarrow f_n\stackrel{L^1}{\rightarrow}f$$

How can we show this, maybe by a subsequence which converges almost everywhere?

I would appreciate if you could help me.

user123234
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2 Answers2

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Hint: Let $\epsilon >0$ (only in the case when measure is finite). Then $$ |f-f_n| = |f-f_n| 1_{ \{|f-f_n| < \epsilon \} } + |f-f_n|1_{ \{ |f_n - f| > \epsilon \} } \leq \epsilon + 2|g|1_{ \{ |f_n - f| > \epsilon \} }$$ Now integrate both sides and use the fact that $f_n$ converge in measure to $f$

In general: Suppose that $f_n$ doesn't converge to f in $L^1$. Then there exist a subsequence $f_{n_k}$ such that for each $k$ $$\int |f_{n_k} - f| d\mu > \epsilon.$$ But from assumptions we know that $f_{n_k}$ converge in measure to $f$, so there exist a subsequence $f_{n_{k_i}}$ of $f_{n_k}$ which converge almost surely to $f$. So from DCT $f_{n_{k_i}}$ converges to $f$. Contradiction

Interpolated
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  • sorry I don't understand this, isn't there an other possibiliy, because I would never have come up with such a hint? I thaught that maybe we can use that there exists a subsequence which converges almost everywhere, and then maybe using the majorised convergence theorem somehow? Don't understand it wrong I appreciate your help, but I would never have come up with such a claim. – user123234 Oct 31 '21 at 16:28
  • Ok. Now I see that my hint works only when measure is finite. – Interpolated Oct 31 '21 at 16:35
  • Shouldn't the subsequence $f_{n_{k_i}}$ converge almost surely to $f$? – user123234 Oct 31 '21 at 18:38
  • Of course it should, typo – Interpolated Oct 31 '21 at 18:49
  • Ah perfect and another last question, is it right that we have a contradiction since $f_{n_{k_i}}$ converges to f only in measure by assumption and not everywhere as we get in the proof? – user123234 Oct 31 '21 at 18:50
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Suppose $f_n\stackrel\mu\to f$. Let $(f_{\varphi(n)})_{n\ge1}$ be any subsequence of $(f_n)_{n\ge1}$. As $f_{\varphi(n)}\stackrel\mu\to f$, there exists a further subsequence $(f_{\varphi(n_k)})_{k\ge1}$ of $(f_{\varphi(n)})_{n\ge1}$ converging to $f$ almost surely. Since all $f_n$'s are dominated by $g$ it follows that $f_{\varphi(n_k)}\to f$ in $\mathrm L^1$. Thus any subsequence of $(f_n)_{n\ge1}$ admits a sub-subsequence converging to $f$ in $\mathrm L^1$. This means that $f_n\to f$ in $\mathrm L^1$.

nejimban
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