How to exactly prove that the square integrable functions on $[0,1]$ are integrable on $[0,1]$? That is, For a function $f$, $\int_{[0,1]}f^2 dx &lt; \infty~\Rightarrow \int_{[0,1]}f dx &lt; \infty$ Or how to prove $\int_{[0,1]}f^2 dx$ diverges whenever $\int_{[0,1]}f dx$ diverges?
122516
I'm reading Real analysis(Stein) and confused by a statement: If $f\in L^2$, then $f\chi_R$ is in both $L^1$ and $L^2$, where $\chi_R$ is the characteristic function of $[-R, R]$.   But $f$ in $L^2$ does not mean that $f\in L^1$. Such an example given by . How could this function on $[-R, R]$ be in $L^1$?
106491
Let $f(x)$ be continuous on [$0,1$] such that $f(x) \ge 0$  for each $x\in[0,1]$.If $\int^{1}_{0}f(x)dx = 0$ , Show that $f(x)=0$ for each $x\in[0,1]$
10801
Show that if $f$ is integrable on $[a, b]$ then $\lvert f \rvert$ is also integrable. Hint: Show that $$U (P , \lvert f \rvert) − L(P , \lvert f \rvert) ≤ U (P , f ) − L(P , f ).$$ I have: $$U (P , \lvert f \rvert) \ge U (P , f )$$ and $$ L(P , \lvert f \rvert) \ge L(P,f)$$ Thus, $$U (P , \lvert f \rvert) − L(P , \lvert f \rvert) ≤ U (P , f ) − L(P , f ).$$
33648
How can we prove the statement?  $f$ is Lebesgue integrable if and only if $|f|$ is Lebesgue integrable.  First, suppose $f$ is Lebesgue integrable, by definition, $f$ is measurable and nonnegative, so $f=|f|$ and $\displaystyle \int_E f \, d \mu=\int_E|f| \, d\mu$ implies that $|f|$ is Lebesgue integrable.   Second, suppose $|f|$ is Lebesgue integrable, if $f$ is nonnegative, then $f=|f|$, so $f$ is Lebesgue integrable; if $f$ is negative, then how can I prove?
107101
Given $X$ is integrable. Prove that $E(X) = \int_{0}^{\infty} [1-F(x)]dx - \int_{-\infty}^{0}F(x)dx$.  My 2ND solution Since $X$ is integrable, $E(X) = E(X^{+}) - E(X^{-}) = \int_{0}^{\infty} xdF_{X^{+}}(x) - \int_{0}^{\infty} xdF_{X^{-}}(x) = \int_{0}^{\infty} [F_{X^{-}}(x) - F_{X^{+}}(x)]dx$.   Now, since $F_{X^{-}}(x) = P(-min(X,0)\leq x) = 1 - P(\min(X,0)&lt;\ -x) = 1 - P(X&lt;-x)$  (since for $x\in (0,\infty), P(0&lt;-x) = 0$)   $= 1 - F_{X}(-x)$, $E[X] = \int_{0}^{\infty} [1 - F_{X}(-x) - F_{X^{+}}(x)] dx = \int_{0}^{\infty} 1 - F_{X}(x) - \int_{0}^{\infty} F_{X}(-x)dx$ (the last equality is because for $x\in (0,\infty), F_{X^{+}}(X) = F_{X}(x)$).   Finally, by using the change of variable $u=-x$ for the 2nd integral, with $u$ as a dummy variable, we easily get the RHS that we need (Q.E.D)  My question: Could someone help review the solution above to see if it's correct this time? Would really appreciate any input.
47468
We suppose that $f : [0, 1] \rightarrow \mathbb{R}$ is a Lipschitz function with constant $K$. We want to show that $f$ is integrable on $[0, 1].$  I've been trying to use the Darboux criterion of integrability by trying to show that, for $\epsilon &gt; 0$, $$S^*(f,P) - S_*(f, P) &lt; \epsilon,$$ where $S^*(f, P)$ and $S_*(f, P)$ are the Upper and Lower Riemann Sums, respectively.  This is what I have so far: $$\sum_{i=1}^n M_i (x_i - x_{i-1})- \sum_{i=1}^n m_i (x_i - x_{i-1}) = \sum_{i=1}^n (M_i - m_i) (x_i - x_{i-1})$$ $$\le \sum_{i=1}^n K (x_i - x_{i-1}).$$ because $|f(x) - f(u)| \le K|x-u| \le K |1|$, since $x\in [0, 1]$. Then, $$\sum_{i=1}^n K (x_i - x_{i-1}) = K.$$ However, this is only a single upper bound, not an arbitrary $\epsilon.$ Can I get some pointers about how to proceed? Thanks!
69661
Let $f(x)$ be continuous on $[a,b]$ and suppose $\int_a^b f(x)g(x)dx = 0$ for every continuous function $g$ on $[a,b]$. Prove that $f(x)=0$ on $[a,b]$.   I understand that $f(x)$ must be zero otherwise the integral can't possibly be zero for all possible $g$. But, I am not sure how to approach this systematically. Any suggestions would be appreciated. Thanks!
109804
Give me an example of two Riemann-integrable functions $f,g:[0,1]\to[0,1]$ such that $g\circ f$ isn't integrable!  I already know the following example: $$f(x)=\begin{cases} 0,     &amp;     \text{if $x$ is irrational} \\ 1,     &amp;     \text{if $x=0$}\\ \frac1q, &amp;     \text{if $x$ is rational and $x=\frac pq$ such that $q\in\Bbb N$ and $(p,q)=1$}  \\ \end{cases}$$ $$g(x) = \begin{cases} 1,  &amp; \text{if $x$ is of the form $\frac 1q$such that $q\in \Bbb N$} \\ 0, &amp; \text{otherwise}  \\ \end{cases}$$ now observe that $g\circ f$ is a famous example of non-integrable function!
45963
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
9297
Let $f : [0, 2] → \mathbb{R}$ be a continuous, positive function and  $$\int_{0}^1 f(x)\,dx = \int_{1}^2 f(x)\,dx= 1$$ For each $x \in [0, 1]$, prove that there exists a unique function $g(x) \in [1, 2]$ such that $$\int_{x}^{g(x)} f(t)\,dt = 1$$ and $g \in C^1$. Well, I think in $g(x) = x +1$, but I'm not sure. Any hint is appreciated.  Thank you.
74641
This is a step in a lecture note I'm reading. It should be simple because the author considers it obvious but I can't see it. What am I missing?  Suppose $U$ and $V$ are integrable over measure space $(\Omega,\mathcal{F},\mu)$. Claim: $$ \int_\Omega\sqrt{U^2+V^2}d\mu\geq\sqrt{\left(\int_\Omega Ud\mu\right)^2+\left(\int_\Omega Vd\mu\right)^2}. $$ Alternative phrasing: with $Z=U+iV$, claim that  $$ \int_\Omega|Z|d\mu\geq\left|\int_\Omega Zd\mu\right|. $$ Thank you for your time!
65071
Let $f:\mathbb{R^+}\to\mathbb{R}$ be an integrable function ($f\in L^1(\mathbb{R}^+,\mathbb{R})$). Do we have $$\lim_{h\to 0^+}\int_0^\infty|f(t+h)-f(t)|dt=0$$ ? How can we prove it ?
43637
Say I have a function $f(x)$ on some interval $[a,b]$.  Say it is integrable such that $\displaystyle\int f(x)~dx $ is defined.  Is $\displaystyle\int f(x)~dx $ necessarily continuous?  If I were to know that the integral is integrable itself such that $\displaystyle\int \int f(x)~dx $ is defined. Would that change anything?   If so why?  Thank you
86068
Let $f_n$ be a sequence of differentaible functions on $[0,1]$ to $\mathbb{R}$ converging uniformly to a function $f$ on $[0,1]$, Then    $f$ is differentiable and Riemann integrable there $f$ is uniformly continuous and R-integrable $f$ is continuous, need not be differentiable on $(0,1)$ and need not be R-integrable on $ [0,1]$, $f$ need not be continuous.   Well I think 2 is only correct statement  as every $f_n$ are differentiable on $[0,1]$ hence uniformly contnous, and as they converge uniformly to $f$ on $[0,1]$ so $f$ is uniformly continuous and every continuous function is rieman integrable..Am I correct?
97444
Suppose $f:[0,1] \to (0,\infty)$ is a Riemann integrable function. Prove that the integral of the function from $0$ to $1$ is strictly positive.   I have been trying to do this for awhile but I can't seem to get it. Here is my thought process: If the function is Riemann integrable, then its set of discontinuities has measure zero (Lebesgue). I am not sure how to connect this to the integral being positive
91883
I want to find a function $f:[0,1] \to [0,1]$ such that $f$ takes on each value in $[0,1]$ exactly twice. I think this means there are an infinite number of discontinuities. Can anyone help me figure this one out?  Anyone have any pointers?
100435
Let $f\in C^2\{[0,1],\mathbb{R})$ and $m\{x\in [0,1]:f(x)=0\}&gt;0$. Prove that $$m\{x\in [0,1]:f'(x)=0\}&gt;0,$$ where $m$ denotes Lebesgue measure.  I don't have any clue to solve this exercise..