Lebesgue integration : Inequality Let $f\in C^1([-2,2],\mathbb{R})$, and $t\in (-1,1)$ and for all $n\geq 1$, $t_n=t+(-1)^n$. I want to prove the following inequality :
$$\Bigg|\int_t^{t_n}f(s)ds\Bigg|\leq c|t_n-t|,$$
where $c=\displaystyle\sup_{x\in[0,2]}|f(x)|$.
My idea using Lebesgue's theory : we can rewrite
$\int_t^{t_n}f(s)ds$ as follows :
$$ \int_{\{s\in\mathbb{R}, \; t\in (t-1,t)\}}f(s)ds+ \int_{\{s\in\mathbb{R}, \; t\in (t,t+1)\}}f(s)ds,$$
which implies that
$$\Bigg|\int_t^{t_n}f(s)ds\Bigg|\leq c(\mu\{s\in\mathbb{R}, \; t\in (t-1,t)\}  + \mu \{s\in\mathbb{R}, \; t\in (t,t+1)\}),$$
but I could not reappear the quantity $|t_n-t|$ again.
 A: If the exercise explicitly says that $c=\sup\{|f(x)| : x\in [0,2]\}$, then the inequality you are looking for is not true. For a counterexample, consider $f:[-2,2] \to \mathbb{R}$ given by $$ f(x)= \begin{cases} x^2, & \text{if} \hspace{2mm} -2\leq x \leq 0, \\
0, & \text{if} \hspace{2mm} 0\leq x \leq 2.
  \end{cases} $$
It is easy to see that $f\in C^{-1}([-2,2],\mathbb{R})$. Now take $t=-\frac{1}{2}$ and consider $t_2=\frac{1}{2}$. Then, on the one hand $$\left| \int_{t}^{t_n} f(x)\ dx \right|=\left| \int_{-\frac{1}{2}}^{\frac{1}{2}} f(x)\ dx \right|= \left| \int_{-\frac{1}{2}}^{0} x^2\ dx \right|=\frac{1}{24}.$$
However, on the other hand, $c=0$ and so $c|t_2-t|=0$
If it was a typo and $c$ is actually equal to $\sup\{|f(x)| : x\in [-2,2]\}$ then the proof is quite easy. Recall that for any integrable function $g:[a,b]\to \mathbb{R}$ the following inequality holds $$\left| \int_{a}^{b} g(x)\ dx \right|\leq  \int_{a}^{b} \left|g(x) \right|\ dx.$$
Having this in mind, notice that for an even $n$ we get $$\left| \int_{t}^{t_n} f(x)\ dx \right| \leq  \int_{t}^{t_n} \left|f(x)\right|\ dx \leq \int_{t}^{t_n} c\ dx = c|t_n-t|.$$
Edit: Adding a separate case to avoid confusion. For an odd $n$ we get
$$\left| \int_{t}^{t_n} f(x)\ dx \right| = \left|- \int_{t_n}^{t} f(x)\ dx \right| =  \left| \int_{t_n}^{t} f(x)\ dx \right| \leq $$ $$ \int_{t_n}^{t} \left|f(x)\right|\ dx \leq \int_{t_n}^{t} c\ dx = c|t_n-t|.$$
Hope this helps.
