# Money account discounted libor rate is it a martingale under risk neutral measure?

I see that Libor $$L(t,S,T)$$ is a martingale under $$T-$$forward measure. Where we used argument that zero-coupon bonds are martingales under $$T$$-forward measure, as zero-coupon bond is a traded security.

We also have arguments that "money account discounted no divident, tradable security is martingale under risk neutral measure". I wonder whether "libor" can be a tradable security.

I find, using $$B(T)$$ the numeraire of risk neutral measure $$\mathbb{Q}$$, $$P(t,T)$$ the zero-coupon bond as numeraire of $$T$$-forward measure, then money account discounted libor:

$$V(t) = B(t)\mathbb{E}^\mathbb{Q}[\frac{L(t,S,T)}{B(T)}]=B(t)\mathbb{E}^\mathbb{T}[\frac{L(t,S,T)}{B(T)}\frac{d\mathbb{Q}}{d\mathbb{T}}]= B(t)\mathbb{E}^\mathbb{T}[\frac{L(t,S,T)}{B(T)}\frac{B(T)P(t,T)}{P(T,T)B(t)} ] = \mathbb{E}^\mathbb{T}[L(t,S,T)P(t,T)]$$

Then by the definition of libor $$L(t,S,T) = \frac{1}{\delta}(\frac{P(t,S)}{P(t,T)}-1)$$, plug in to the last term above, we get

$$V(t) = \frac{1}{\delta}\mathbb{E}^{T}[P(t,S)-P(t,T)] =\frac{1}{\delta}\mathbb{E}^{T}[\frac{P(t,S)-P(t,T)}{P(T,T)}] = \frac{1}{\delta} \frac{P(t,S) - P(t,T)}{P(t,T)}$$

Where I put $$P(T,T)$$ in the denominator as we know it's value is 1.

then we got $$V(t) = \frac{L(t,S,T)}{B(t)} = \mathbb{E}^\mathbb{Q}[\frac{L(t,S,T)}{B(T)}]$$ ? I feel there are some issue somewhere and would like to know how people explain it...

So what's the reason that libor cannot be treated as a tradable asset ? When can we add $$P(T,T)=1$$ in the denominator ?