Why is the Hazard function not a pdf? I am trying to understand why the hazard function is not a PDF. For a random variable T, people often define the PDF of this random variable as: 
$$f(t)=\lim_{\delta \to 0} \frac{P(t\leqslant T <t+\delta)}{\delta}$$
By this definition, the hazard should also be a conditional PDF.
$$\lambda(t)=\lim_{\delta \to 0} \frac{P(t\leqslant T < t+\delta \mid T\geqslant t)}{\delta}$$
Seems like the two functions share the same type of definition! They are the limit of a probability. So why is one a PDF and the other not?
I guess is the reason this is not a PDF because the conditioning is not on a single event $T=t$ but rather on $T\geqslant$? If it were on a single event $T=t$, would this be a PDF?
 A: I think the counterexample Björn suggested is a enough to answer the question. Let me write it out in more detail:
Björn's Counterexample
Let $T\stackrel{d}{=}\mathrm{Exp}(\lambda)$ for any $\lambda > 0$ then $f_T(t) = \lambda e^{-\lambda t}$ which implies $S(t) = \mathcal{P}(T>t) = 1-(1-e^{-\lambda t}) = e^{-\lambda t}$.
The hazard function can be calculated as $\lambda(t) = \dfrac{f_T(t)}{S(t)} = \dfrac{\lambda e^{-\lambda t}}{e^{-\lambda t}} = \lambda$. (hence the notation $\lambda(t)$ for a hazard function).
Clearly this implies that $\lambda(t)$ is not a probability density function, since $\int_0^{+\infty} \lambda(u) \mathrm{d}u \not = 1$.
Back to your question(s)
The first part of your question was: 

`Why is one a pdf and the other not'

As far as I remember, $f_T(t)$ is defined as a pdf if certain conditions hold. One of these conditions is $\int_\Omega f_T(t) \mathrm{d} t =1$. (Which can clearly be violated in the case of $\lambda(t)$).
The density function $f_T(t)$ can be viewed as the limit of a cumulative distribution function in a narrow timeslot, which is what the notation $f_T(t) = \lim_{\delta \to 0} \frac{\mathcal{P}(t\leqslant T<t+\delta)}{\delta}$ suggests. I would not call this the definition since it lacks the conditions.

I guess is the reason this is not a pdf because the conditioning is not on a single event $T=t$ but rather on $T\geqslant t$? If it were on a single event $T=t$, would this be a pdf?

Notice how:
$$
 \mathcal{P}(t\leqslant T<t+\delta \mid T = t) = 1
$$
which implies that conditioning on $T=t$ isn't that usefull.
A conditional probability function $f_{Y\mid X}(y\mid X=x)$ for continious r.v. is defined as
$$
f_{Y\mid X}(y\mid X=x) = \dfrac{f_{Y,X}(Y=y,X=x)}{f_X(x)}
$$
and the hazard function does not fit here. You need (at least) two different random variables. You could look at something like this:
$$
f_{T \mid Z}(t\mid Z = 1)
$$
for a certain $Z$ as an indicator of treatment or control. Then cleary this is a probability function.
A: The argument of a conditional pdf cannot depend on the conditioning event in any way, shape or form.  In 
$$f_{T\mid A}(t\mid A) = \lim_{\delta\to 0} \frac{P\{t < T \leq t+\delta\mid A\}}{\delta},$$ $A$ can be a fixed event such as $\{T>5\}$ but not something that depends on $t$ such as $\{T > t\}$.
Another important reason why a hazard function $h(t)$ (or any scalar submultiple thereof) cannot possibly be a pdf is that
$$\int_0^\infty h(t)\, \mathrm dt =  \infty$$ whereas pdfs of lifetimes have more mundane values for their integrals: 
$$\int_0^\infty f_T(t)\, \mathrm dt =  1.$$
A: The Hazard function can be equivalently seen as the ratio of the pdf $f(t)$ to the survival function $S(t)$:
$$\lambda(t)=\lim_{\delta \to 0} \frac{P(t\leqslant T < t+\delta \mid T\geqslant t)}{\delta}$$
by definition of conditional probability:
$$=\lim_{\delta \to 0} \frac{P(t\leqslant T < t+\delta \cap T\geqslant t) / P(T \ge t)}{\delta}$$
$$=\lim_{\delta \to 0} \frac{P(t\leqslant T < t+\delta) / P(T \ge t)}{\delta}$$
by definition of $f(t):$
$$= \frac{f(t)}{P(T \ge t)}$$
by definition of survival function:
$$ =\frac{f(t)}{S(t)}$$
Thus since the hazard function is the pdf scaled by the survival function, it will not generally integrate to $1$, and hence is not a pdf.
