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Let $(f_n)^{\infty}_{n=1}$ be a sequence of measurable (not-necessarily $\ge 0$). Let $g \gt 0$ be a measurable function with $\int g d\mu < \infty$ (integrable) such that $f_n\ge -g$ a.e. relative to $\mu$ in $E\in S$ ($S$ $\sigma$-algebra).

I want to show that $$\int_E \liminf f_n d\mu \le \liminf \int_E f_n d\mu.$$

My attempt: Let $N_n=\{x\in E; f_n(x) < g(x)\}$. Then $\mu(N_n)=0$ for all $n \in \mathbb{N}$ by hypothesis. Let $N= \cup N_n$, it follows that $N$ is also $\mu$-null.

With $N$ as above, one can see that $$(f_n+g)\chi_{E \setminus N}\ge 0,$$ everywhere, so applying Fatou's Lemma one gets: $$ \int_{E\setminus N} \liminf (f_n+g) d\mu \le \liminf \int_{E\setminus N} (f_n +g) d\mu.$$

Since $\liminf g =g, \liminf \int g= \int g,$ and $\int g < \infty$, if I could "open" both integrals as the sum of the integrals of each function, I could cancel out the terms involving $g$, yet I can't seem to make it work. I have a semi-linearity theorem for positive measurable functions and a linearity theorem for integrable functions, but the $f_n$'s are neither positive nor integrable.

I was told that I should prove a linearity theorem concerning non positive measurable functions whose integrals can be extended real numbers, i.e., we might define $$\int f = \int f^+ -\int f^-,$$ given that said difference is well-defined (we do not get $\infty - \infty$).

I can't seem so see the light in the proof of said linearity. Any insight on the issue would be greatly appreciated.

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You can apply Fatou's lemma to $h_n = f_n + g \geq 0$. Since $\liminf h_n = \liminf f_n + g$, it yields, $$ \int (\liminf f_n + g)\,d\mu \leq \liminf \int (f_n + g)\,d\mu $$ Conclude using the condition $\int g\,d\mu < \infty$

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  • $\begingroup$ That is the line of attack outlined above. The problem is that since $f_n$ is not positive nor integrable (finite integral) I can't open up integrals of sums as sums of integrals (I don't have a linearity condition in this case, this is what I am trying to get). $\endgroup$ – alonso s Oct 10 '13 at 7:56
  • $\begingroup$ Notice that $f_n$ is integrable iff $(f_n+g)$ is integrable and consider the different cases. $\endgroup$ – Siméon Oct 10 '13 at 8:13
  • $\begingroup$ How does one see that $f_n$ is integrable iff $(f_n+g)$ is? $\endgroup$ – alonso s Oct 10 '13 at 9:15
  • $\begingroup$ One has $|f_n| - |g| \leq |f_n + g| \leq |f_n| + |g|$ by triangle inequality. $\endgroup$ – Siméon Oct 10 '13 at 12:43
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Here is the detail solution for your question:

Apply Fatou's Lemma for $f_n + g \geq 0,$ we obtain

$$\int_{E}\liminf_{n\longrightarrow+\infty}( f_n +g)\leq \liminf_{n\longrightarrow+\infty} \int_E (f_n+g) $$

The idea is to expand both sides.

For the LHS

Let $E^+ =\{x \in E\,:\, \liminf\limits_{n \longrightarrow +\infty}f_n(x) \geq 0\}$ and $E^- =\{x \in E\,:\, \liminf\limits_{n \longrightarrow +\infty}f_n(x) <0\}$, we rewrite the LHS of first inequality $$ \int_{E}\liminf_{n\longrightarrow+\infty}( f_n +g) = \int_E \left[(\chi_{E^+})\liminf_{n\longrightarrow+\infty}( f_n +g) + (\chi_{E^-})\liminf_{n\longrightarrow+\infty}( f_n +g) \right]\\ = \int_E (\chi_{E^+})\liminf_{n\longrightarrow+\infty}( f_n +g) + \int_E(\chi_{E^-})\liminf_{n\longrightarrow+\infty}( f_n +g) \\ = \int_{E^+}\liminf_{n\longrightarrow+\infty}( f_n +g) + \int_{E^-}\liminf_{n\longrightarrow+\infty}( f_n +g) \\ = \int_{E^+}(\liminf_{n\longrightarrow+\infty} f_n +g) + \int_{E^-}(\liminf_{n\longrightarrow+\infty}f_n +g)\qquad(A) \\ = \int_{E^+}\liminf_{n\longrightarrow+\infty} f_n +\int_{E^+}g + \int_{E^-}\liminf_{n\longrightarrow+\infty}f_n +\int_{E^-}g \qquad (B) \\ = \int_{E}\liminf_{n\longrightarrow+\infty} f_n +\int_{E}g.$$

We have $(A)=(B)$ because $\liminf\limits_{n\longrightarrow+\infty} f_n $ and $g$ are nonnegative on $E^+$ and they are integrable on $E^-$ (from $f_n \geq -g$, we have $\liminf\limits_{n\longrightarrow+\infty} f_n \geq -g$, hence on $E^-$, $0 \geq\liminf\limits_{n\longrightarrow+\infty} f_n \geq -g$).

For the RHS

The main point is to prove $$ \int_E (f_{n} + g) = \int_{E} f_{n} + \int_E g,\qquad \forall n \in \mathbb{N}. \qquad (C) $$ (both sides can be equal to infinity).

First, if $\int_E |f_n| < \infty$, then the identity is true because both functions are integrable. In the other case, that means $\int_E |f_n| = \infty$, put $f_n^+ = \max\{f_n,0\}$ and $f_n^-=-\min\{f_n,0\}$, then
$$ \int_E |f_n |= \int_E f_n^+ + \int_E f_n^- $$ (because $|f_n|=f_n^+ +f_n^-$ and $f_n^+$, $f_n^-$ are nonnegative, the above identity follows from Theorem 1.27 p.22, Rudin, Real and Complex Analysis.)

Moreover, $ 0 \leq f_n^- \leq g$ so $f_n^-$ is integrable. It follows that $\int_E f_n^+ = \infty.$ Therefore, $$ \int_E f_n = \int_E f_n^+ - \int_E f_n^- = \infty. $$ (see Definition 1.31, identity (2), p.25, Rudin, Real and Complex Analysis.)

Hence $$\int_{E} f_{n} + \int_E g = \infty.$$ We need to prove $$\int_E (f_{n} + g) = \infty.$$ Assume by contradiction that $\int_E (f_{n} + g) < \infty.$ Since $|f_n| \leq |g| + |f_n +g| = |g| + (f_n +g)$ (because $f_n + g \geq 0$), we have

$$\int_E |f_n| \leq \int_{E} \left[ |g| + (f_n +g) \right] = \int_{E} |g| + \int_{E} (f_n+g) < +\infty $$ which is a contradiction and thus $(C)$ is proved.

The RHS of first inequality can be rewritten as $$ \liminf_{n\longrightarrow+\infty} \int_E (f_n+g) = \liminf_{n\longrightarrow+\infty} \left(\int_E f_n+\int_E g\right) = \liminf_{n\longrightarrow+\infty} \int_E f_n + \int_E g $$

The result follows.

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    $\begingroup$ Why "$\forall N \exists n_N ; \int f_{n_N}=+\infty$" implies "$\liminf\int f_n = +\infty$"? $\endgroup$ – Filburt Jan 18 '17 at 23:57
  • $\begingroup$ @FernandoVieiraCostaJúnior It is really a serious mistake. Thank you so much for pointing out. I have corrected my answer. $\endgroup$ – macnguyen Jan 19 '17 at 16:29

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