I'm working on a nice problem in measure theory. Given $\mathbb{R}$ with its standard Lebesgue measure on its $\sigma$-algebra and a Lebesgue integrable function $f: \mathbb{R} \rightarrow \mathbb{C}$, what could we say about the limit $$\lim_{t\rightarrow 0} \int |f(x)-f(x+t)|d\lambda(x)$$ Of course with continuous functions, this limit is always equal to zero. But I can see some difficulties arrising quickly. Maybe one can force this integral to be nonzero somewhere when taking a suitable limit for $t$.   Certainly when trying to make this into a rigorous proof (that the limit is equal to zero), I did not see a clear way to get this result.  I'm not looking for a complete solution but any hints would be appreciated!
35848
I am having a hard time with the following real analysis qual problem.  Any help would be awesome.  Suppose that $f \in L^p(\mathbb{R})$, where $1\leq p&lt; + \infty$. Let $T_r(f)(t)=f(t−r)$. Show that $\lim_{r \to 0} \|T_rf−f\|_{L_p} =0$, that is $$ \lim_{r\to 0} \left( \int_{\mathbb R} |f(t-r) - f(t)|^p \mathrm d t\right)^{1/p} =0.$$
87218
Suppose that $f$ is a continuous and real function on $[0,\infty]$. How can we show that if $\lim_{n\rightarrow\infty}(f(na))=0$ for all $a&gt;0$ then $\lim_{x\rightarrow+\infty} f(x)=0$?
11285
Assume that $f$ be continuous on $\mathbb{R}$, $f'(x)$ exists for all $x\neq 0$, and $\lim_{x\rightarrow 0} f'(x)=1$. We need to show $f'(0)$ exist and is equal to $1$.  $f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$, $\lim_{x\rightarrow 0}f'(x)=1\Rightarrow\lim_{x\rightarrow 0}\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=1$...am I going in the right direction? Please help.
58485
I'm learning about measure theory, specifically the Lebesgue integral of nonnegative functions, and need help with the following problem.       Let $f:\mathbb{R}\to[0,\infty)$ be measurable and $f\in L^1$. Show that $F(x)=\int_{-\infty}^x f$ is continuous.   I know is isn't much but the only thing a could think of is that given $x, y \in \mathbb{R}$ with $x &lt; y$ we note that $F(x) \leq F(y)$, i.e. $F$ is increasing. So maybe we can apply one of the convergence theorems of Lebesgue integration theory.    I was also wondering if this problem can be solved using only Riemann integration theory.
17835
Let $(X,B,\mu)$ be a complete measure space,Show that $$\lim _{q \rightarrow \infty}\|f\|_{q}=\|f\|_{\infty}, \quad \forall f \in \bigcup_{p} \bigcap_{p \leqslant q&lt;\infty} L^{q}$$ So,$\lim _{q \rightarrow \infty}\|f\|_{q}$ , $\|f\|_{\infty}$ are equal-norm with space $ L^{\infty} \cap\left(\bigcup_{p} \bigcap_{p \leqslant q} L^{q}\right)$. Case 1: $m(X)&lt;\infty $.It's easy to prove that. Case 2: $m(X)=\infty $. I have no idea about it,And I started to doubt the correctness of this conclusion. Can somebody give me a hint for this problem or just give an example to prove that this is a wrong conclusion when $m(X)=\infty $. Thanks in advance.
101469
Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous on $\mathbb{R}$ and that $f(r)=0$ for all $r \in \mathbb{Q}$. Prove that $f(x)=0$ for all $x \in \mathbb{R}$.   My attempt: Define a sequence $(x_n)$ where $x_n \in \mathbb{Q}$ for all $n \in \mathbb{N}$ and assume that $(x_n) \rightarrow a \not\in \mathbb{Q}$. Since $f$ is continuous, we have $\lim_n{f(x_n)}=f(a)=0$. Since $a$ is arbitrary irrational number, we have $f(a)=0$ for all $a \not\in \mathbb{Q}$. Hence, we proved the statement.  Is my proof valid? or is there any flaw ?
57620
Let $f: [1, \infty) \to \mathbb{R}$ be a continuous function such that the improper integral $$\int_1^\infty f(x) \ dx$$ exists. Show or disprove that $\lim \limits _{x \to \infty} f(x) =0$.  Our professor said that there are counterexamples but after frying my brain out I still could not find any. It does sound logical.
122687
Let $f:[0, 1] \rightarrow \mathbb{R}$ a continuous function. If $a&gt;0$, show that:  $$ f(0)\ln(\frac{b}{a})=\lim_{\epsilon\rightarrow 0}\int_{\epsilon a}^{\epsilon b} \frac{f(x)}{x}dx$$  Tried using Riemann sum, but did not succeed.
38333
Assume $f$ is Lebesgue measurable and bounded on $[0,1]$ and that $$\int_{0}^{1}\frac{|f(x+h)-f(x)|}{h}\;dx\to0\;\text{as}\;h\to0.$$ Show that $f$ is constant for almost every $x$ in $[0,1]$.  My question relates to my proposed proof below.  At no point did I rely on the boundedness of $f$ on $[0,1]$, only its measurability and so I suspect my reasoning is faulty and would appreciate somebody that could (in their answer) scrutinize and correct my proof.  (Proposed) Proof. Assume first that $h\to0^{+}$ and suppose that we may interchange the two limits.  Then since the difference quotients are non-negative, we conclude $f'(x)$ exists and is equal to $0$ for almost every $x$ in $[0,1]$, and the claim follows. (As pointed out, we don't necessarily know the limit of the integrands exists even when we assume the interchange is valid, although this is probably not difficult to reason one's way out of -- more damning is below.)  The goal quickly turns to justifying this interchange.  To that end, the presumed existence of the above limit implies that for any $0&lt;M&lt;\infty$ there is some positive real number $h(M)$ such that for any $h&lt;h(M)$ we have $$\left|\int_{0}^{1}\frac{|f(x+h)-f(x)|}{h}\;dx\right|&lt;M.$$ Since $m([0,1])=1$ (in particular, we are integrating over a set of finite measure) and the difference quotients are non-negative, we conclude that $$\frac{|f(x+h)-f(x)|}{h}&lt;M\;\text{a.e.}\;x.$$ Since this bound is independent of $h$ once $h&lt;h(M)$, it follows that at some point along the sequence each of the difference quotients is pointwise a.e. dominated by $M$, which is integrable on any set of finite measure.  Hence the interchange of limits discussed above is valid, and with the case of $h\to0^{-}$ being handled similarly, the claim is proved.   (As the comment points out, we cannot conclude anything about the upper (or lower) bound of a function based off the value of its integral, even if the integration is taken over a set of finite measure and the integrand in question is nonnegative; approximations to the identity serve or otherwise spikey functions on sets of small measure as counter-examples.)    Edit. Fatou's lemma implies $$\int_{0}^{1}\liminf_{h\to0}\frac{|f(x+h)-f(x)|}{h}\;dx\leq\lim_{h\to0}\int_{0}^{1}\frac{|f(x+h)-f(x)|}{h}\;dx=0.$$  Hence $$\liminf_{h\to0}\frac{|f(x+h)-f(x)|}{h}=0\;\text{a.e.}\;x.$$ (The $\liminf$ of course exists without further justification.)  We now need a way to obtain the same for $\limsup_{h\to0}\frac{|f(x+h)-f(x)|}{h}.$  This could be accomplished by showing the limit exists under the assumptions given, in which case the $\liminf$ is equal to the $\limsup$.  Any ideas how to execute this, or otherwise approach the problem?  I'm wondering too if my original argument above could be modified by taking into account the assumed boundedness of $f$ in order to exclude the possibility that its difference quotients form a "spikey" sequence of integrands.    Edit. Here is another approach I am considering by posting this question   Peter Tamaroff mentioned the type-writer sequence being a counter-example to the post linked.
75574
I have some problems understanding  of this problem:  Given $f,g : \mathbb{R} \rightarrow \mathbb{R}$ - continuous functions with period $1$, prove that  $$\lim _{n \rightarrow \infty} \int_0^1 f(x)g(nx) \, dx = \int_0^1 f(x) \, dx \cdot \int_0^1g(x) \, dx$$  It says there that if we split this integral into $n$ parts to get  $$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} \int_{\frac{r}{n}}^{\frac{r+1}{n}} f(x)g(nx) \, dx$$  we will have   $$\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right) \int_{\frac{r}{n}}^{\frac{r+1}{n}}g(nx) \, dx$$  because for large $n$, $f$ is roughly constant over the range.  Could you tell me why $f$ is roughly constant over the range for $n$ large enough?  I would really appreciate all your insight.
33602
My question is the following :   is there a $\sigma$-algebra $\mathcal{T}$ (of subsets of $\mathbb{R^n}$) that contains strictly the $\sigma$-algebra $\mathcal{L}$ of  Lebesgue measurable sets (in $\mathbb{R}^n$), and such that there is a measure on $\mathcal{T}$ that extends the usual Lebesgue measure on $\mathcal{L}$ ?  I guess not, but I did not find a reference.
73949
Let $f: \mathbb R \rightarrow \mathbb R $ be a continuous function such that $\int_{0}^{\infty} \,f(x) dx$ exists.  Then Prove that incase   (i) $f$ is a non negative function, then $\lim_{x\rightarrow \infty} f(x) $ must exist and is $0$.  (ii) $f$ is a positive differentiable function , $\lim_{x\rightarrow \infty} f'(x) $ must exist and is $0$  $Attempt$: For the first part, i don't have a rigorous proof except for the fact that the given condition can be visualised geometrically. Since, the definite integral is actually calculating the area beneath the non negative function, the only way the given limit can exist when limit of f(x) itself tends to 0 at infinity.  Please give me a direction so that i can make this proof rigorous enough.  For the second part, i took an example. We know that ( leaving out the finite integration parts from $0$ to $1$ ..) $\int_{1}^{\infty} e^{-x^2} dx \leq \int_{1}^{\infty} e^{-x} dx$ and the latter converges.  But the derivative of $e^{-x^2} = (-2x)e^{-x^2}$ whose integration does not exist when x $\in~[1,\infty)$ as it's a monotonic function after a finite $x$.  Any help in providing rigor to the above proof will be very helpful  Thanks
82533
In the book of Analysis on Manifolds by Munkress, at page 121, it is asked that     Let $A$ be a bounded open set in $\mathbb{R}^n $; let $f: \mathbb{R}^n &gt; \to \mathbb{R} $ be a bounded continuous function. Give an example   where $\int_{\bar A} f$ exists but $\int_A f$ does not.   I couldn't find any such example. I mean I do noticed that $A$ should not be rectifiable, and the only way I can think of that $f$ is not integrable on $A$ is that the set $E$ where the limit  $$\lim_{x \to x_0} f(x) = 0$$ fails for $x_0 \in \partial A$ has a non-zero measure, but then this set $E$ also contained in $\bar A$, so in such a case $\int_{\bar A} f$ also wouldn't exists.  In short, I'm looking for such an example.  Note: I have read  answer, but I have no idea what the answerer is talking about, so I wanted to ask a new question including my own thoughts.
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Let $g:[1, \infty)\rightarrow \mathbb R$ a continuous non-negative function, such that $\int_{1}^{\infty} g(x)\ dx$ converges. Is it true that $\lim\limits_{x\to \infty}g(x)=0$ ?   I tried to find a counter-example but I can't figure a trivial one. I also tried to prove it by the definition of the convergence of $g(x)$ but couldn't show that limit is really $0$.
106583
Let $(X,\mathcal{M},\mu)$ be a measure space and $f$ be a nonnegative measurable function on $X$. Let $1\le p&lt;\infty$. Show that, the function $|f|^p$ is integrable with respect to $\mu$ precisely when the function $$t\mapsto t^{p-1}\mu({x:|f(x)&gt;t})$$ is integrable on $[0,\infty)$ with respect to Lebesgue measure. In addition prove that    $$\int_X |f|^p=p\int^{\infty}_{0} t^{p-1}\mu({x: |f(x)&gt;t}) dt\,$$   I solved this problem for the case when space is sigma finite but don't know how to solve for general case.
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1) Construct a continuous function $f$ on $\mathbb{R}$ that is integrable on $\mathbb{R}$ but $\displaystyle\limsup_{x \to \infty} f(x)  = \infty$.  I took the function that is equal to $n$ on $[n, n + 1/n^{3})$ and made it continuous by saying that $f$ is the line segment joining $n$ and $n+1$ on $[n + 1/n^{3}, n+1)$. But I am failing to prove this integrable. For this, $\lim f(x) = \infty$ but how do you prove in general, if limit does not exist that $\limsup$ is infinity?  2) Prove that if $f$ is uniformly continuous and integrable on $\mathbb{R}$ we have $\displaystyle\lim_{|x| \to \infty} f(x) = 0$.  Any help is appreciated.  Thanks
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Suppose $\{f_n\} \subset L^+$ , $f_n \rightarrow f$ pointwise, and $\int f = \lim \int f_n &lt; \infty$. Then $\int_{E} f = \lim \int_{E} f_n$ for all $E \in \mathcal{M}$. However, this needn't be true if $\int f = \lim \int f_n = \infty$.   This is the exercise 13 on page 52, Real Analysis, by Gerald B. Folland. $L^+$ is the space of all measurable functions from $(X,\mathcal{M}, \mu)$ to $[0, \infty]$.  I can find a counterexample satisfying $\int f = \lim \int f_n = \infty$. Let $X = \Bbb N$, $\mathcal{M} = \mathcal{P}(\Bbb N)$ and $\mu$ be the counting measure. $f_n = \chi_{\{2i+1:i \leq n\} \cup \{2j:n \leq j \leq 2n \}}$. Then $\int f = \lim \int f_n = \infty$. But when $E = \{2k: k \in \Bbb N\}$,  Then $\int_{E} f =  0$ and $\lim \int_{E} f_n = \infty$.  But how to show the part "$f_n \rightarrow f$ pointwise, and $\int f = \lim \int f_n &lt; \infty$. Then $\int_{E} f = \lim \int_{E} f_n$ for all $E \in \mathcal{M}$"?