Let $\alpha$ be nondecreasing on $[0,1]$. Then Riemann-Stieltjes integrable function is well defined with respect to the integrator $\alpha$. I was wondering if the following assertion is true:  (1) If $f$ is Riemann-Stieltjes integrable on $[0,1],$ then $f$ is Lebesgue integrable on $[0,1]$ with respect to the measure induced by $\alpha$. (2) Do we need additional hypotheses on $\alpha$ for guaranteeing the above assertion? For example, $\alpha$ is right-continuous.  I would be grateful if you give any comment for my question.
120982
I thought every Riemann integrable function is lebesgue integrable. But , a user comments that the statement is not correct.  I'm confused.  Can someone clarify?
115455
Is is true that if a function is Riemann integrable, then it is Lebesgue integrable with the same value? If it's true, how to prove it? If it's false, what is a counterexample?
112274
I have been given the example $$ f_n(x)= \begin{cases} 1, &amp; \text{if}\; x\in\{(x_k)_{k=1}^{n}\}, \\ 0, &amp; \text{otherwise.} \end{cases} $$ Here the sequence $(x_k)_{k=1}^{\infty}$ is the sequence that "enumerates" elements of $\mathbb{Q}$. The sequence of functions converges pointwise to $$ f(x)= \begin{cases} 1, &amp; x\in\mathbb{Q}, \\ 0, &amp; x\notin\mathbb{Q}, \end{cases} $$ which isn't Riemann integrable. However, I am not quite convinced that $f_n$ is integrable. I would like some help in seeing how $f_n$ is Riemann integrable. Also, I would like more examples of sequences of Riemann integrable functions $f_n$  that converge pointwise to a function $f$ that is not Riemann integrable.
63284
Ok, so one of the formulations of the Riemann-Lebesgue Lemma says: $$ f\in L^1(\mathbb{R}) \implies \hat{f}(\omega)\to 0\;\mbox{ when } \;|\omega|\to\infty.$$  I get all the steps of the proof, except the one which says that if $f$ is an arbitrary integrable function, it may be approximated in the $L^1$ norm by a compactly supported smooth function $g\in L^1$. My question is: Why? How do you know that for all $\epsilon&gt;0$, and for all arbitrary function $f\in L^1$ exists a compactly supported smooth function $g$ so that $||f-g||_{L^1}&lt;\epsilon$? The "proof" can be found in many places, but this step can be found in Wikipedia
75012
I have to decice if the following function is Lebesgue-integrable on $[0,1]$:  $$g(x)=\frac{1}x\cos\left(\frac{1}x\right) $$ where $x\in[0,1]$.
19019
For an arbitrary interval $I$, how can we find a positive on $I$ integrable function? And how does one construct a measurable but not integrable function. If not all measurable functions are integrable, what is the best way to determine if a function is L-Integrable?   thx in advance
108752
I am working on the following problem from Stein and Shakarchi:  Let $f$ be an integral function on $\mathbb{R}^d$ such that $\|f\|_{L^1} = 1$ and let $f^*$ by the Hardy-Littlewood maximal function corresponding of $f$.     Prove that if $f$ is integrable on $\mathbb{R}^d$, and $f$ is not identically zero, then $f^*(x) \geq c/|x|^d $ for some $c &gt; 0$ and all $|x| &gt; 1$.   Conclude that $f^*$ is not integrable on $\mathbb{R}^d$.   Then, show that the weak type estimate $m(\{x:f^*(x) &gt; \alpha \}) \geq c/\alpha$ for all $\alpha &gt; 0$ whenever $\int|f| = 1$, is best possible in the following sense: if $f$ is supported in the unit ball with $\int|f| = 1$, then $m(\{x:f^*(x) &gt; \alpha \}) \geq c'/\alpha$ for some $c' &gt; 0$ and all sufficiently small $\alpha$.     [Hint: For the first part, use the fact that $\int_B|f| &gt; 0$ for some ball $B$.]  I am not really sure where to begin with this.  Any help would be greatly appreciated.
92682
If $f,g$ are Riemann integrable on $[a,b]$, and $f(x) &lt; g(x)$ for all $x \in [a,b]$, prove that $$ \int_a^b f(x) \,dx &lt; \int_a^b g(x) \,dx$$  This is a strict inequality. I know how to prove the monotonicity of integrals with the non strict inequality, but I am not sure how to do this.
113682
Let $f:[0,1]\rightarrow \mathbb R$  be  bounded and  integrable  on  $(\delta,1]$ . Is  $f$  integrable  on  $[0,1]$ $?$   I  thought  of  the  function  ${1}\over  {x}$  which  could  satisfy  the   condition  and  prove  in  negative  only  it  is  not  bounded  on  $[0,1]$  ,  in  fact  not  even  defined  at  $0$ .  I'm unsure where to go from here.  Thanks.
107794
Prove for bounded $f$, if $f:[a,r] \to \mathbb{R}$ is Riemann-Integrable for $r \in [a,b)$ then $f:[a.b] \to \mathbb{R}$ is Riemann-Integrable and $\displaystyle \lim_{r \to b, r &lt; b} \int_a^r f= \int_a^bf$.  Alternative definition for Riemann-Integrable functions :: $f : [a,b] \to \mathbb{R}$ is Riemann-Integrable if $\forall \varepsilon &gt; 0, \exists$ step functions $f_\varepsilon, g_\varepsilon$ such that $\lvert f- f_\varepsilon \rvert \le g_\varepsilon$ and $Ig_\varepsilon \le \varepsilon$ where $Ig_\varepsilon = \sum_{j = 0}^{N}g_\varepsilon(x)(\Delta x_{j+1} - \Delta x_j)$.  $f$ is bounded so $\displaystyle \exists M \in \mathbb{R}^+$ s.t. $|f(x)| \le M, \forall x \in [a,b] $ so we know  $\displaystyle|\int_a^r f| \le \int_a^r|f| \le M(r-a) \le M(b-a) $. Now, for some index set $I$ where $r_0 \le r_1 \le \ldots\le r $ we have $\displaystyle \int_a^rf = \sum_{i \in I}\int_{a}^{r_{i}}f$  Do I need to show $f$ is Riemann-Integrable over $[a,b]$ before I can solve the limit? It seems like it but seems like the other way would give it to me.  EDIT:  Since $f(x) \le M, \forall x \in [a,b]$ and we have there exists $r \in [a,b]$ such that $b - \frac{1}{n+1} &lt; r &lt; b$ we have $\displaystyle \int_a^{b-\frac{1}{n+1}}|f| \le \int_a^r|f| \le M(b-a)$ Does this give me that $f$ is Riemann-Integrable over $[a,b]$?
18402
Let $f$ integrable and let $E_\alpha=\{x\mid |f(x)|&gt;\alpha\}$ for $\alpha\geq 0$. Show that $$\int\left|f(x)\right|dx=\int_0^\infty m(E_\alpha)d\alpha$$ where $m$ is the Lebesgue measure.   By hypothesis, $f$ is measurable, therefore there exists a sequence of simple function such that $\varphi_n\nearrow |f|.$ Moreover, $E_\alpha^n\nearrow E^\alpha$. By the continuity of the measure, $m(E_\alpha^n)\to m(E_\alpha)$ for all $\alpha\geq 0$. By the convergence monotone theorem, $$\int_0^\infty  m(E_\alpha)d\alpha=\lim_{n\to\infty }\int_0^\infty  m(E_\alpha^n)\underset{(*)}{=}\lim_{n\to\infty }\int\varphi_n(x)dx=\int\left|f(x)\right|dx.$$   I don't understand the equality $(*)$, why $m(E^n_\alpha)=\varphi_n(x)$ ?
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
90992
Theorem $:$ Let $f : [a,b] \longrightarrow \Bbb R$ be a Riemann integrable function. Then $f \in L_1[a,b]$ and $$\int f\ d\lambda = \int_{a}^{b} f(x)\ dx.$$  The proof given in my book is as follows $:$     At the fag end of the proof $f$ has been proved to be measurable which uses the completeness of the measure space $\left ([a,b], \mathcal L \cap [a,b], \lambda |_{[a,b]} \right ).$ Then it has been stated that $f \leq \Phi_n,$ $\forall$ $n$ and hence $f \in L_1[a,b]$ since $\Phi_n \in L_1[a,b]$ by the proposition $5.4.3 \ $(i) which states the following $:$  Let $f \in \Bbb L$ and $g \in L_1[a,b].$ If $|f(x)| \leq g(x)$ for a.e. $x(\mu).$ Then $f \in L_1[a,b].$  I don't know how to use this proposition to claim that $f \in L_1[a,b].$ Also I don't understand the way dominated convergence theorem used here. What I know about dominated convergence theorem is as follows $:$  Let $(X, \mathcal S, \mu)$ be a complete measure space. Let $g \in L_1(X)$ and $\{f_n \}_{n \geq 1}$ be a sequence in $\Bbb L$ such that $|f_n(x)| \leq g(x)$ a.e. $x(\mu).$ Let $f_n(x) \to f(x)$ as $n \to \infty$ a.e. $x (\mu).$ Then $f \in L_1(X)$ and $$\int f\ d\mu = \lim\limits_{n \to \infty} \int f_n\ d\mu.$$  From here how to conclude that $$\Psi_n\ \bigg\uparrow\ f \implies \int f\ d\lambda = \lim\limits_{n \to \infty} \int \Psi_n\ d\lambda.$$ Please help me in this regard. Thanks in advance.
88681
Prove that if $f:[a,b] \rightarrow \mathbb{R}$ is integrable, then so is $|f|$ and $\left|\int_{a}^{b}f(x)dx\right| \leq \int_{a}^{b}|f(x)|dx$   I already proved that |f| is integrable, but how do I show that $|\int_{a}^{b}f(x)dx| \leq \int_{a}^{b}|f(x)|dx$   ?
59902
If we have a real valued function $f$ continuous at some point $a$, is it necessarily true that $f$ is Riemann integrable on the interval $[a - \delta, a + \delta]$ if a $\delta &gt; 0$ exists?
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?
53690
Suppose $f$ is differentiable everywhere on $[0,1]$. Must $f$ be absolutely continuous on $[0,1]$? I know this is true if $f'$ is integrable but I'm not sure in this more general case.