PROBLEM: 1) Assume $f$ is a bounded measurable function on $\mathbb{R}$, does $f_n(x):=f(x+1/n)$ converge to $f(x)$ almost everywhere? 2) If not, is there a subsequence $n_k$ such that $f_{n_k}$ converges to $f$ almost everywhere? Beginner in Real Analysis and have two ideas but neither worked:  Lusin's Theorem. Not sure how the statement would be if the domain of $f$ is $\mathbb{R}$. One related problem here: . Dirichlet Function as an example. When $x$ is irrational, whether $f_n(x)$ converges to $f(x)$ or not.  Any hints or counterexamples will be appreciated.
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We know that if $f\in L^1(\mathbb{R})$, then $\|f(\cdot+1/n)-f(\cdot)\|_{L^1}\to 0$ as $n\to \infty$, which implies that there exists a subsequence $f_{n_k}=f(x+\frac{1}{n_k})$ such that $f_{n_k}\to f$ a.e.   My problem is that  Is there any $f\in L^1(\mathbb{R})$  function such that the almost convergence result  $$ f(x)=\lim_{n\to\infty}f(x+1/n)\quad \text{a.e.} $$  is NOT valid?   I intend to believe that there exits such function. Otherwise, there would be a beautiful statement such that $f(x)=\lim_{n\to\infty}f(x+1/n)$ a.e. for any $f\in L^1(\mathbb{R})$.   Thank you for the long commenting below @David C. Ullrich. This question has been reduce to find a compact set $K$ such that for any $x\in K$, there are infinite $n$ such that $x+\frac{1}{n}\not\in K$.
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Hi can you help me solve this exercise? Thanks. Let $f: [0;+\infty) \to \mathbb{R}$ be nonnegative and continuous function. Suppose $\forall a &gt; 0$  $\sum_{n=1}^{\infty} f(na)$ is convergent. Prove that $\int_{0}^{\infty}f(x) dx$ is convergent. I tried to solve it by using the Riemann sum, but for fixed a it doesn't work. I have no other ideas.
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Given a sequence $f_n \in L^p$ and $g \in L^p$, with $|f_n| \leq g$, I am trying to show that $f_n \to f$ in measure implies $f_n \to f$ in $L^p$.  Firstly, I know that if $f_n \to f$ in measure, then there is a subsequence $f_{n_i}$ such that $f_{n_i} \to f$ almost everywhere. Then I can use the dominated convergence theorem to show that $\lVert f_{n_i} - f_p\rVert \to 0$.  Now I am trying to show that $\lVert f_n - f\rVert_p \to 0$. My idea is to assume that $\lVert f_n - f\rVert_p \nrightarrow 0$ and then show that this contradicts the fact that $\lVert f_{n_i} - f\rVert_p \to 0$, but I am not sure of the details. Can anyone help me finish the argument?
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My question is related to:   and   Let $f\colon (0, \infty)\to R$ be  continuous such that  $f(x)\leq f(nx)$ for all positive $x$ and natural $n$.  It was proved that the limit (finite or infinite) in the infinity exists. Do we know if such a function must be (weakly) increasing? I believe that there might be counterexamples.
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Let $f : \mathbb R^n \rightarrow \mathbb R$ be a measurable function.  We know that $f$ is not necessarily locally integrable - but how hard can a measurable $f$ fail to be so? For example, is it true that $f$ is locally integrable outside of a negligible set $N \subset \mathbb R^n$?   Edit: In consideration of the comment by zhs, let me correct the phrasing of the question. I was originally wondering whether the 'singular' behaviour of a measurable function can be localized to a 'small' set.   A motivational example is $f(x) = 1/x$ over the real line. It is locally integrable in the complement of the origin. Hence, a conceivable phenomenon, which I had in mind originally when asking the question, could be that a measurable $f$ is integrable over open sets that are compactly contained in the complement of some negligible sets.
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Assume that $f_n → f$ uniformly on $S$ and each $f_n$ is continuous on $S$. Let $(x_n)$ be a sequence of points in $S$ converging to $x \in S$. Then $f_n(x_n) → f(x)$.  I'm stuck in thinking about it and I don't manage to prove it neither to find a counterexample. Thanks for your help.
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I am trying to show that for nonnegative $f$ on $\mathbb{R}$ if $||f||_1&lt;\infty$, we have  $$ \lim_{p\uparrow \infty}||f||_p=||f||_\infty $$ I have tried to fiddle around with  $$ \left(\int_\mathbb{R}|f|^p|f|^{1-p} \right)^{1/p}&lt;\infty $$ but haven't found a nice way to get an estimate on $\int_E|f|^p$ from this. Honestly I am pretty uncomfortable with non finite measure spaces.   Any hints would be awesome! Any references for problems similar to this would also be awesome.  edit: this is not a duplicate of the linked question as 1) my measure space is not finite, 2) my function is in $L^1$ and not necessarily in $L^\infty$
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Suppose $f$ is a continuous function at $x = c$ in $[a,b]$.  Prove that for any sequence ${x_n}$ in $[a,b]$ converting to $c$, the sequence $\{f(x_n)\}$ converges to $f(c)$.  That is, $$ \lim_{n\to\infty}f(x_n)= f\left(\lim_{n\to\infty}x_n\right)$$    This proof seems simple but there are a few things that I need to know first. If $\{x_n\}$ converges to $c$, is it sufficient to substitute $c$ in for $\lim_{n\to\infty}x_n$?  Also needing some guidance on the structure of this proof.  Thanks!
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I'm sure I'm having a notational misunderstanding. Anyway, suppose $(f_n)$ is a sequence of continuous functions from a metric space $X$ to $\mathbb{R}$. So, if $(f_n)$ converges uniformly to a function $f$, then  $$\lim_{n \to \infty} \inf_{x \in X} f_n(x) = \inf_{x \in X} \lim_{n \to \infty} f_n(x)$$  and  $$\lim_{n \to \infty} \sup_{x \in X} f_n(x) = \sup_{x \in X} \lim_{n \to \infty} f_n(x).$$  Can somebody explain to me why is that? Again, this might be really simple...
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Problem Statement:  Let $\{f_n\}$ converge uniformly on a set $E$. Suppose that each $f_n$ is bounded. Prove that there exists an $M &gt; 0$ s.t. $|f_n(x)| \leq M$ for all $x \in E$ and all $n = 1,2, ...$.  Proof:  If each $f_n$ is bounded by some $M_n &gt; 0$ then can't we just take $M = \text{max}\{M_n\}$?  I don't see what's so difficult here. Am I not using the fact that $\{f_n\}$ converges uniformly?
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Let $n$ be a positive integer.  $\;  $  Let $f : \mathbb{R}^n \to \mathbb{R}^n$ be everywhere .  Let $S$ be a subset of $\mathbb{R}^n$ with  zero.  Does it follow that $\{f(s) : s\in S\}$ has Lebesgue measure zero?  (I know it would if $D(f)$ is continuous.)
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Suppose that $f$ is uniformly continuous on $\mathbb{R}$. If $\displaystyle \lim_{n\to\infty} y_n =0$ and $f_n(x) := f(x + y_n)$ for $x \in \mathbb{R}$, prove that $f_n$ converges uniformly on $\mathbb{R}$.
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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.
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Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.  I know that by the Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence. But I need some hints as to how to prove the question above. Thanks for your help!
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Monotone Convergence Theorem for general measure: Let $(X,\Sigma,\mu)$ be a measure space. Let $f_1, f_2, ...$ be a pointwise non-decreasing sequence of $[0, \infty]$-valued $\Sigma-$measurable functions, i.e. for every $k\ge 1$ and every $x$ in $X$, $$0 \le f_k(x) \le f_{k+1}(x).$$ Next, set the pointwise limit of the sequence ${f_n}$ to be $f$. That is, for every $x$ in $X$, $$f(x) = \lim_{k\to \infty}f_k(x).$$ Then $f$ is $\Sigma-$measurable and $$\lim_{k\to \infty}\int f_k d\mu = \int f d\mu.$$  I've noticed that when it comes to monotone convergence theorem (either Lebesgue or general measure), usually its definition restricts the monotone function sequences to be nonnegative. I'm not sure why the 'nonnegative' is necessary.
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Let $\{f_n\}$ be a sequence of analytic functions on a region $\Omega$ with $|f_n| ≤ 1$ on $\Omega$. Let K be compact and contained in $\Omega$. Suppose $\{f_n\}$ converges at infinitely many points in K. Then is it true or false that $\{f_n\}$ necessarily converges at every point of $\Omega$?
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Please help me out with this problem:  Let $f_n(x) : [0,1] \to \mathbb R$ be a sequence of continuous functions convergent at every $x \in [0,1]$ to a continuous function $f: [0,1]  \to \mathbb R$ . Does $f_n$ converge uniformly to $f$?  My first solution: I read this example in  posted earlier and I think it works: Take a sequence $f_n:[0,1] \to \mathbb R$ such that $f_n$ increases linearly from $0$ to $1$ on the interval $\left[ 0,\frac{1}{n} \right]$, decreases linearly from $1$ to $0$ on the intreval $\left[ \frac{1}{n}, \frac{2}{n} \right]$, and is $0$ on $\left[ \frac{2}{n},1 \right]$. Then $f_{n} \to 0$ nonuniformly.  Is this answer true? Also, if anyone can give some examples of such function with detailed proof, I will very appreciative?