# Monotone and Dominated convergence Theorem implies L1-convergence

Dominated Convergence Theorem: Let $$f_n$$ be measurable and assume that $$f_n$$ converges a.e. to $$f$$. If $$|f_n(x)|\leq g(x)$$ for some integrable $$g$$ it follows that $$f_n$$ converges in $$L^1$$ to $$f$$, in particular $$\int f_n\to\int f$$. Hence, we do not only have the convergence of the integrals but even $$L^1$$-convergence.

Now my question: Does the monotone convergence theorem also implies $$L^1$$ convergence or just $$\int f_n\to\int f$$?

• Hint $$\int_{\mathbb R}|f_n-f|=\int_{\mathbb R}f-\int_{\mathbb R}f_n.$$ – Surb Sep 29 '18 at 10:26
• So the answer must be yes since the right side converges to 0 by MCT. In the literature it is often noticed that DCT provide $L^1$-convergence, but never for MCT. So I thought that this will not hold in general. – RandomUser Sep 29 '18 at 10:32
• The problem is that $\int f$ can by infinite. So, indeed MCT do not provide $L^1$ convergence. But if $\int f<\infty$, then $f_n\to f$ in $L^1$. – Surb Sep 29 '18 at 11:27
• Thank you, that makes sense. – RandomUser Sep 29 '18 at 11:45