If $|\int f_n-\int f |\to 0 \Rightarrow\int |f_n - f| \to 0$.?

Question

If $|\int f_n -\int f| \to 0 \Leftrightarrow |\int (f_n-f) |\to 0$ where $f_n$ and $f$ have the hypothesis of dominated convergence theorem

It is true that:

If $|\int f_n-\int f |\to 0 \Rightarrow\int |f_n - f| \to 0$.?

Answer 1

If both integrals are well-defined (both $f$ and $g$ are integrable), $$\left\lvert \int (g-f) \right\rvert \leq\int \lvert g-f \rvert$$ by the triangle inequality.

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Edit: to answer the (new) question: no, the other implication is false. Look at the example given above by Solomonoff: you need to assume some convergence of $f_n$ to $f$ to hope that such implication holds.

Edit 2: also, to clarify: what do you mean by "Comply with the hypothesis of dominated convergence theorem"?

Answer 2

This statement is false. For example, let $f = 0$ and $f_n = \text{id} \cdot \chi_{[-1, 1]}$, the identity function on $[-1, 1]$. Then $\int f_n - f = 0$ so $|\int f_n - f| \rightarrow 0$ but $\int |f_n-f| = 1$. The opposite implication is true by Clement C's post.

Answer 3

There seems to be a problem with the question as it is currently stated since $\int \left|f_n - f\right| \to 0$ is a consequence of the hypotheses of the dominated convergence theorem.

I think the OP's question is this:

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Assume $f_n$ and $f$ satisfy most of the hypotheses of the dominated convergence theorem, i.e. $|f_n|\leq|g|$ for some integrable function $g$ and all $f_n$ are measureable, but that $f_n$ doesn't converge to $f$ pointwise.

If $f_n\rightarrow f$ pointwise, assuming the rest of the hypotheses of the dominated convergence theorem hold, we get that $$\int \left|f_n - f\right| \to 0$$ which also implies that $$ \left|\int f_n -\int f\right| \to 0.$$

The question:

If we do not assume $f_n\rightarrow f$ pointwise, but instead assume that $$\left|\int f_n-\int f \right|\to 0,$$ is it true that $$\int \left|f_n - f\right| \to 0?$$

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In general $\left|\int f_n-\int f \right|\to 0$ does not tell you much about the functions except that the have the same area underneath them in the limit. As noted by "@Salomonoff's Secret", $f_n(x)=x$ on $[-1,1]$ and $f(x)=0$ have the same integral, but $\int \left|f_n - f\right|=1$.