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Let $f$ a non-negative Lebesgue integrable function $ ( f\in L^+([0 ,\infty)) )$ such that $\displaystyle{ \lim_{x \to +\infty} f(x)}$ exists (finite or infinity). Prove that $\displaystyle{\lim_{x \to +\infty} f(x) =0}$.

Here it is the only thing I did.

Consider an increasing sequence $ (f)_n$ of simple non-negative functions such that $ f_n \to f$ pointwise. Then $ \int f =\lim \int f_n$.

1st case: $\displaystyle{\lim_{x \to +\infty} f(x) = l < +\infty}$.

Let $ \epsilon >0$, then there exists $ r>0 $ such that $ |f(x)-l|< \epsilon , \quad \forall x>r$.

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  • $\begingroup$ On a related note, it's a good exercise to construct an integrable function for which $\lim_{x \to +\infty} f(x)$ does not exist. $\endgroup$ – Nate Eldredge Feb 2 '12 at 14:08
  • $\begingroup$ @NateEldredge: $\displaystyle{ f:= \chi_{\mathbb{Q}}}$ is an integrable function , $\displaystyle{ \int f=0}$ and $\displaystyle{\lim_{x \to \infty} f(x)}$ does not exist. $\endgroup$ – passenger Feb 2 '12 at 14:11
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    $\begingroup$ Yep, that works. Now can you find one which is continuous? $\endgroup$ – Nate Eldredge Feb 2 '12 at 14:33
  • $\begingroup$ @NateEldredge: This is more difficult... I have to think about it $\endgroup$ – passenger Feb 2 '12 at 14:42
  • $\begingroup$ Hint to Nates exercise: 1) Start to think about a convergent series $A= \sum a_n$. 2) One way to construct a function $f$ such that $\int f = A$ is to put $f(x)=a_n$ on the interval $I_n=(n,n+1)$. Can you find an other way? (Note that $|I_n|=1$, fairly thick...?) $\endgroup$ – AD. Feb 2 '12 at 16:10
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Argue by contradiction: There are only two other possible cases. Suppose $ \displaystyle \lim_{x\to \infty} f(x) = L $ where $ L \in (0,\infty) .$ Then by definition of a limit, there exists $ a \in \mathbb{R}^+ $ such that $ f(x) > L/2 $ for all $ x> a.$ Estimating the integral: $$ \int^{\infty}_0 f(x) dx = \int^a_0 f(x) + \int^{\infty}_a f(x) dx > \int^a_0 f(x) dx + \frac{L}{2} \int^{\infty}_a 1 dx = \infty$$

which contradicts the assumption that $f\in L^1.$ Now you try to construct a similar argument for the case $L=\infty.$

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  • $\begingroup$ Thank's for the answer! For the case $L=\infty$ I said that for all $\epsilon>0$ $ \exists r >0 : f(x) > 1/\epsilon , \forall x\geq r$ and continue the same as you. I have one question. In your solution where do you use properties of Lebesgue integral (as for example did in beginning ) I mean this solution uses only properties of Riemann integral. $\endgroup$ – passenger Feb 2 '12 at 14:20
  • $\begingroup$ I used only basic properties that both the Riemann and Lebesgue integrals share. All Riemann integrals are also Lebesgue integrals anyway. $\endgroup$ – Ragib Zaman Feb 2 '12 at 14:25
  • $\begingroup$ Yes I know this. So it does not need to use the definition of Lebesgue integrable functions? $\endgroup$ – passenger Feb 2 '12 at 14:27
  • $\begingroup$ We need the condition that $f \in L^1$ for the integrals in the proof to even make sense, and then we invoke it at the end for the contradiction, but not so much other than that. $\endgroup$ – Ragib Zaman Feb 2 '12 at 14:33

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