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Is it true that if $(f_n)_{n\geq 1}$ is a sequence of Lebesgue-integrable functions on $[0,1]$ that converges pointwise almost everywhere to a function $f$, then $f$ is integrable?

Edit: If not, then can we at least say something about the convergence of $(f_n)_{n\geq 1}$ in $L^1$, that is, is it true that $\|f-f_n\|_1\to 0$, when $n\to\infty$?

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No. Look at the monotone covergence theorem and dominated convergence theorem to find sufficient conditions for this to hold. –  Michael Greinecker Oct 22 '12 at 17:41

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No this is not true. For example, consider the sequence $$f_k(x)=\chi_{(1/k,1]}(x)\cdot x^{-1}.$$ Then $f_k(x)\rightarrow f(x)=1/x$ for a.e. $x\in[0,1]$ and the $f_k$ are clearly in $L^1([0,1])$, while $f$ is not (note that here $x=0$ is not a problem since elements of $L^1$ are by definition only given up to a set of measure zero).

However, if there exists an integrable function $g$ such that $|f_k(x)|\le g(x)$ for a.e. $x\in[0,1]$ and every $k$, then the dominated convergence theorem (the sequence is dominated by $g$, hence the name) says that also the pointwise limit $f$ is integrable and $$\lim_{k\rightarrow\infty}\int^1_0 f_k(x) dx=\int^1_0 f(x)dx.$$ Note that the last equation is equivalent to saying that $$\int^1_0 |f_k(x)-f(x)| dx\longrightarrow 0\;\text{as}\;k\rightarrow\infty,$$ which is just saying that $||f_k-f||_1\rightarrow 0$, i.e. $f_k$ converges to $f$ in the $L^1$ norm.

Regarding your last question: We can also not conclude convergence in $L^1$ in general! This follows from the example above because $L^1$ is complete, i.e. $f_k\in L^1$ and $f_k\rightarrow f$ in $L^1$ implies $f\in L^1$.

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