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Let $(X,F,\mu)$ be finite measure space.

Let $f_n$ be sequence of measurable function from $X$ and $f_n\geq 0$ almost everywhere $\mu$.

Claim.

If $\lim_{n\to 0}$$\int_X f_n d\mu=0$, then $f_n$ converges to $0$ a.e $\mu$?

Intutively It's true.

Can you help me?

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    The claim is actually false. You can even take $f_n$ to be characteristic functions of smaller and smaller sets which wander wildly over $X$. For an explicit counter-example, take Example 4 from these notes: https://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/ – 2017-01-20

2 Answers 2

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Intuition may betray you. Here is an example:

Consider $(X, \mathcal{F}, \mu) = ([0, 1), \mathcal{B}([0,1)), \mathrm{Leb})$ be the unit interval equipped with the Borel $\sigma$-algebra and the Lebesgue measure restricted to $[0, 1]$. Consider the double sequence

$$ g_{n,k}(x) = \mathbf{1}_{[k/2^n, (k+1)/2^n)}(x) = \begin{cases} 1, & \text{if }\frac{k}{2^n} \leq x < \frac{k+1}{2^n} \\ 0, & \text{otherwise} \end{cases} $$

where $n \geq 0$ and $k = 0, 1, \cdots, 2^{n-1}$. Now re-enumerate them to form a sequence $(f_n)$:

$$ (f_1, f_2, f_3, \cdots) = (g_{0,0}, g_{1,0}, g_{1,1}, g_{2,0}, \cdots, g_{n,k}, g_{n,k+1}, \cdots, g_{n,2^n-1}, g_{n+1,0}, \cdots). $$

Then it follows that $\int_{X} f_n \, d\mu \to 0$ as $n\to\infty$, but for every $x \in [0, 1)$ there are infinitely many $n$ for which $f_n(x) = 1$ and infinitely many $n$ for which $f_n(x) = 0$. Therefore $f_n$ converges nowhere.

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A consequence of the statement, if it were true, would be that $L^1$ convergence implies a.e. $\mu$ convergence in your measure space.

Indeed, assume that $f_n$ converges in $L^1$ to $f$. By definition, this means that $ \int_{X} |f - f_n| \, d\mu \to 0$. Since $|f - f_n| \geq 0$, the statement, if it were true, would imply that $|f-f_n| \to 0$ a.e. $\mu$.

Since $L^1$ convergence does not imply a.e. $\mu$ convergence, the claim is false.