0
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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: terrytao.wordpress.com/2010/10/02/… $\endgroup$ – Josh Keneda Jan 20 '17 at 11:27
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.