If $f$ is integrable and bounded over $[\delta,1]$ for all $\delta \in (0,1]$ then is $f$ integrable over all of $[0,1]$ $?$ 
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.
 A: Let $M$ such that $|f(x)|\leq M$ for all $x\in [0,1]$. Let $F(x)=\int_x^1 |f(x)|dx$ with $0<x<1$. The function $F$ is nonincreasing and bounded since 
$$0\leq F(x)\leq M\int_0^1dx=M,$$
therefore it converges.
A: The answer is positive: if $f$ is integrable over $[\delta,1]$ for all $\delta \in (0,1]$ then $f$ is integrable over $[0,1]$.
Following is a proof using Riemann integration.
Take $\epsilon > 0$. By hypothesis $f$ is bounded on $[0,1]$, so you can find $M > 0$ with $\vert f(x) \vert \le M$ for $x \in [0,1]$. Take $\delta= \frac{\epsilon}{4M}$. As $\delta > 0$ and $f$ is supposed to be integrable on $[\delta,1]$, you can find two step functions $s,S$ defined on $[\delta,1]$ with $$s(x) \le f(x) \le S(x) \text{ and } \int_\delta^1 (S(x)-s(x)) dx \le \frac{\epsilon}{2}$$
Then extends $s,S$ on $[0,1]$ by defining $s(x)=-M$ and $S(x)=+M$ for $x \in [0,\delta)$. You have $s(x) \le f(x) \le S(x)$ for $x \in [0,1]$, $s,S$ are steps functions and $$\int_0^1 (S(x)-s(x)) dx = \int_0^\delta (S(x)-s(x)) dx + \int_\delta^1 (S(x)-s(x)) dx \le \frac{\epsilon}{2} +\frac{\epsilon}{2}=\epsilon.$$
As this is true for all $\epsilon > 0$, you can conclude that $f$ is Riemmann (hence Lebesgue) integrable on $[0,1]$.
A: If $f$ is integrable in $[\frac{1}{n},1]$ for every $n$, so the set of descontinuities $C_n$, on $[\frac{1}{n},1]$ has zero measure. Therefore, the set of descontinuities on $[0,1]$ has zero measure and since $f$ is bounded on $[0,1]$ you have that $f$ is integrable.
