I am trying to prove Lemma 4.1 (1) from Olav Kallenberg's Foundations of Modern Probability:
$\mu\{x:|x|>r\}\le\tfrac r2\int_{-2/r}^{2/r}(1-\varphi(t))\,dt$
From context, I believe that $\varphi$ is a characteristc function, and $\mu$ is a probability measure. The proof given by my lecturer (which differs slightly from that in Kallenberg) starts out okay:
$\begin{align}\int_{-c}^c(1-\varphi(t))\,dt &= \int_{-c}^c\left(1-\int\exp(itx)\,\mu(dx)\right)\,dt\\ &=\int\mu(dx)\int_{-c}^c(1-\exp(itx))\,dt\\ &=2c\int \left(1-\frac{\sin(cx)}{cx}\right)\,\mu(dx)\tag{1} \end{align}$
Recall that $\exp(iz)=\cos z+i\sin z$, so $$\int_{-c}^c \exp(itx)\, dt=\frac{\exp(icx)-\exp(-icx)}{ix}=\frac{2\sin(cx)}{x}.$$ If $|cx|>2$, then $\frac{\sin(cx)}{cx}<1/2$, implying that $\left|1-\frac{\sin(cx)}{cx}\right|\ge \frac{1}{2}.$
I vaguely follow everything going on up until this point, we use Tonelli and then remark on the integral at hand, but the conclusion drawn is something that I do not see at all:
So, $(1)\ge c\mu\{x:|cx|>2\}$. Let $c=2/r$ to obtain the result.
What happns to the case when $|cx|\le 2$? I am not sure how we get from the remarks to conclude that "$(1)\ge c\mu\{x:|cx|>2\}$"?
For the case $|cx|<2$, the result follows by showing that the integral is non-negative, as per stocasticboy's comment.
To see this, by sketching the graph, it should be obvious that for $|t|<2$, the function $\tfrac 1 t\sin t$ is non-negative (see sketch on Wolfram Alpha).